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## Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

### Samacheer Kalvi 11th Chemistry Gaseous State Textual Evaluation Solved

I. Choose the correct answer from the following:

Question 1.

Gases deviate from ideal behavior at high pressure. Which of the following statement (s) is correct for non – ideality?

(a) at high pressure the collision between the gas molecule become enormous

(b) at high pressure the gas molecules move only in one direction

(c) at high pressure, the volume of gas become insignificant

(d) at high pressure the inter molecular interactions become significant

Answer:

(d) at high pressure the inter molecular interactions become significant

Question 2.

Rate of diffusion of a gas is …………

(a) directly proportional to its density

(b) directly proportional to its molecular weight

(c) directly proportional to its square root of its molecular weight

(d) inversely proportional to the square root of its molecular weight

Answer:

(d) inversely proportional to the square root of its molecular weight

Question 3.

Which of the following is the correct expression for the equation of state of van der Waals gas?

Answer:

Question 4.

When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules ………….

(a) are above inversion temperature

(b) exert no attractive forces on each other

(c) do work equal to the loss in kinetic energy

(d) colide without loss of energy

Answer:

(b) exert no attractive forces on each other

Question 5.

Equal weights of methane and oxygen is mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen ………..

(a) 1/3

(b) 1/2

(c) 2/3

(d) 1/3 × 273 × 298

Answer:

(a) 1/3

Hint:

mass of methane = mass of oxygen = a

number of moles of methane = \(\frac {a}{16}\)

number of moles of Oxygen = \(\frac {a}{32}\)

mole fraction of Oxygen =

Partial pressure of oxygen = mole fraction x Total Pressure = \(\frac {1}{3}\)P

For an ideal gas, the partial pressure formula relationship is called Henry’s Law.

Question 6.

The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called …………

(a) Critical temperature

(b) Boyle temperature

(c) Inversion temperature

(d) Reduced temperature

Answer:

(b) Boyle temperature

Hint:

The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called Boyle temperature

Question 7.

In a closed room of 1000 m^{3} a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?

(a) Viscosity

(b) Density

(c) Diffusion

(d) None

Answer:

(c) Diffusion

Question 8.

A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be ………….

(a) At the center of the tube

(b) Near the hydrogen chloride bottle

(c) Near the ammonia bottle

(d) Throughout the length of the tube

Answer:

(b) Near the hydrogen chloride bottle

Hint:

Rate of diffusion α 1/√m

m_{NH3 } = 17

m_{HCl} = 36.5

γ_{NH3 } > γ_{HCl}

Hence white fumes first formed near hydrogen chloride.

Question 9.

The value of universal gas constant depends upon ………..

(a) Temperature of the gas

(b) Volume of the gas

(c) Number of moles of the gas

(d) units of Pressure and volume

Answer:

(d) units of Pressure and volume

Question 10.

The value of the gas constant R is …………

(a) 0.082 dm^{3} atm.

(b) 0.987 cal mol^{-1} K^{-1}

(c) 8.3 J mol^{-1}K^{-1}

(d) 8 erg mol^{-1}K^{-1}

Answer:

(c) 8.3 J mol^{-1}K^{-1}

Question 11.

Use of hot air balloon in sports at meteorological observation is an application of

(a) Boyle’s law

(b) Newton’s law

(c) Kelvin’s law

(d) Brown’s law

Answer:

(a) Boyle’s law

Question 12.

The table indicates the value of van der Waals constant ‘a’ in (dm^{3})^{2} atm. mol^{-2}

The gas which can be most easily liquefied is ………….

(a) O_{2}

(b) N_{2}

(c) NH_{3}

(d) CH_{4}

Answer:

(c) NH_{3}

Hint:

Higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquefaction. Option (c) is correct

Question 13.

Consider the following statements.

(i) Atmospheric pressure is less at the top of a mountain than at sea level

(ii) Gases are much more compressible than solids or liquids

(iii) When the atmospheric pressure increases the height of the mercury column rises Select the correct statement.

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (i) and (iii)

(d) (i), (ii) and (iii)

Answer:

(d) (i), (ii) and (iii)

Question 14.

Compressibility factor for CO_{2} at 400 K and 71.0 bar is 0.8697. The molar volume of CO_{2} under these conditions is ………..

(a) 22.04 dm^{3}

(b) 2.24 dm^{3}

(c) 0.41 dm^{3}

(d) 19.5 dm^{3}

Answer:

(c) 0.41 dm^{3}

Compressibility factor (z) = \(\frac {Pv}{nRT}\)

V = \(\frac {z x nRT }{p}\)

V = 0.41 dm^{3}

Question 15.

If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes ………….

(a) 4P

(b) 2P

(c) P

(d) 3P

Answer:

(c) P

Hint:

P_{2} = P_{2} Option (c)

Question 16.

At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3\(\sqrt{3}\) times that of a hydrocarbon having molecular formula What is the value of n?

(a) 8

(b) 4

(c) 3

(d) 1

Answer:

(b) 4.

Hint:

Squaring on both sides and rearranging

27 x 2 = m_{Cn}H_{2n-2}

54 = n(12) + (2n-2)(l)

54 = 12n+2n – 2

54 = 14n – 2

n = (54 + 2)/14 = 56/14 = 4

Question 17.

Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape. (NEET phase 1)

(a) 3/8

(b) 1/2

(c) 1/8

(d) 1/4

Answer:

(c) 1/8

Hint:

The fraction of oxygen that escapes in the time required for one half of the hydrogen to escape is 1/8

Question 18

The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\)For an ideal gas α is equal to ………..

(a) T

(b) 1/T

(c) P

(d) none of these

Answer:

(b) 1/T

Hint:

Question 19.

Four gases P, Q, R and S have almost same values of ‘b’ but their ‘a’ values (a, b are Van der Waals Constants) are in the order Q < R < S < P. At a particular temperature, among the four gases the most easily liquefiable one is ………….

(a) P

(b) Q

(c) R

(d) S

Answer:

(a) P

Hint:

Greater the ‘a’ value, casier the liquefaction

Question 20.

Maximum deviation from ideal gas is expected from (NEET)

(a) CH_{4(g)}

(b) NH_{3(g)}

(c) H_{2(g)}

(d) N_{2(g)}

Answer:

(b) NH_{3(g)}

Question 21.

The units of Van der Waals constants ‘b’ and ‘a’ respectively

(a) mol L^{-1} and L atm^{2} mol^{-1}

(b) mol L and L atm mol^{2}

(c) mol^{-1} L and L^{2} atm mol^{-1}

(d) none of these

Answer:

(c) mol^{-1} L and L^{2} atm mol^{-1}

Hint:

an^{2}/V^{2} atm

a = atm L^{2}/mol^{2} = L^{2} mol^{-2} atm

nb = L

b = L /mol = L mol^{-1}

Question 22.

Assertion : Critical temperature of CO_{2} is 304 K, it can be liquefied above 304 K.

Reason : For a given mass of gas, volume is to directly proportional to pressure at constant temperature.

(a) both assertion and reason arc true and reason is the correct explanation of assertion

(b) both assertion and reason are true but reason is not the correct explanation of assertion

(c) assertion is true but reason is false

(d) both assertion and reason are false

Answer:

(d) both assertion and reason are false

Hint:

Correct Statement: Critical temperature of CO_{2} is 304 K. It means that CO_{2} cannot be liquefied above 304 K, whatever the pressure may applied. Pressure is inversely proportional to volume.

Question 23.

What is the density of N, gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K^{-1} mol^{-1})

(a) 1.40 g/L

(b) 2.81 g/L

(c) 3.41 g/L

(d) 0.29 g/L

Answer:

(c) 3.41 g/L

Hint:

Density = \(\frac {Mass}{Volume}\)

Question 24.

Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas ? (T is measured in K)

Answer:

For a fixed mass of an ideal gas V α T

P α 1/V

and PV = Constant

Question 25.

25 g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?

(a) HBr

(b) HCl

(c) HF

(d) HI

Answer:

(d) HI

Hint:

At a given temperature and pressure

Volume α number of moles

Volume α Mass / Molar mass

Volume α 28 / Molar mass

i.e. if molar mass is more , volume is less. Hence Hl has the least volume.

II. Answer these questions briefly.

Question 26.

State Boyle’s law.

Answer:

At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.

Mathematically, Boyle’s law can be written as

V ∝ \(\frac{1}{P}\) …………..(1)

(T and n are fixed, T-temperature, n-number of moles)

V = k × \(\frac{1}{P}\) …………(2)

k – proportionality constant

PV = k (at constant temperature and mass)

Question 27.

A balloon filled with air at room temperature and cooled to a much lower temperalure can be used as a model for Charles’ law.

Answer:

Charles’ law:

- V T at constant P and n (or) \(\frac {V}{T}\) = Constant
- A balloon filled with air at room temperature and cooled to a much lower temperature. the size of the balloon is reduced. Because if the temperature of the gas decreases, the volume also decreases in a direct proportion.
- When temperature is reduced, the gas molecules inside in over slower due to decreased temperature and hence the volume decreases.

Question 28.

Name two items that can serve as a model for Gay Lussac’s law and explain.

Answer:

Gay Lussac’s law:

1. P α T at constant volume (or) = \(\frac {V}{T}\)

2. Example – 1:

You fill the car type completely full of air on the hottest day of summer. The type cannot change it shape and volume. But when winter comes, the pressure inside the lyre is reduced and the shape is also reduced. This confirms that pressure and temperature are direct related to each other.

3. Example – 2:

The egg in the bottle experiment.

A glass bottle is taken, inside the bottle put some pieces of cotton with fire. Then place a boiled egg (shell removed) at the top of the bottle. The temperature inside the bottle increases from the fire, rising (he pressure. By scaling the bottle with egg, the fire goes on, dropping the temperature and pressure. This causes the egg to be sucked into the bottle.

P α T is proved (or) = \(\frac{P_{1}}{V_{1}}=\frac{P_{2}}{V_{2}}\)

Question 29.

Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means.

Answer:

The mathematical expression that relates gas volume and moles is Avogadro’s hypothesis. It may be expressed as

V ∝ n,

\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = constant

where V_{1} and n_{1} are the volume and number of moles of a gas and V_{2} and n_{2} are a different set of values of volume and number of moles of the same gas at the same temperature and pressure.

Question 30.

What are ideal gases? In what way real gases differ from ideal gases.

Answer:

- Ideal gases are the gases that obey gas laws or gas equation PV = nRT.
- Real gases do not obey gas equation. PV = nRT.
- The deviation of real gases from ideal behaviour is measure in terms of a ratio of PV to nRT. This is termed as compression factor (Z). Z = \(\frac {PV}{nRT}\)
- For ideal gases Z = 1.
- For real gases Z > 1 or Z < 1. For example, at high pressure real gases have Z >1 and at intermediate pressure Z < 1.
- Above the Boyle point Z> 1 for real gases and below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with the increase in pressure.
- So, it is clear that at low pressure and high temperature, the real gases behave as ideal gases.

Question 31.

Can a Van der Waals gas with a = 0 be liquefied? Explain.

Answer:

If the van der Waals constant (a) = 0 for gas, then it behaves ideally, (i.e.,) there is no intermolecular forces of attraction. So it cannot be liquefied. Moreover,

P_{c} = \(\frac{a}{27 b^{2}}\)

If a = 0, then P_{c} = 0; therefore it cannot he liquefied.

Question 32.

Suppose there ¡s a tiny sticky area on the wan of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?

Answer:

- Molecules hitting the tiny sticky area on the wall of the container of gas moves faster as they get closer to adhesive surface, but this effect is not permanent.
- The pressure on the sticky wall is greater than on the ordinary area of walls.

Question 33.

Explain the following observations?

(a) Aerated water bottles are kept under water during summer

(b) Liquid ammonia bottle is cooled before opening the seal

(c) The type of an automobile is inflated to slightly lesser pressure in summer than in winter

(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude

Answer:

(a) Aerated water bottles contains excess dissolved oxygen and minerals which dissolved under certain pressure and if this pressure suddenly decrease due to change in atmospheric pressure, the bottle will certainly burst with decrease the amount of dissolved oxygen in water, this tends to change the aerated water into normal water.

(b) The vapour pressure of ammonia at room temperature is very high and hence the ammonia will evaporate unless the vapour pressure is decreased. On cooling the vapour pressure decreases so that the liquid remains in the same state. Hence, the bottle is cooled ’ before opening.

(c) In Summer due to hot weather conditions, the air inside the tyre expands to large volumes due to heat as compared to winter, therefore, inflated to lesser pressure in summer.

(d) As we move to higher altitude, the atmospheric pressure decreases, therefore the balloon can J easily expand to large volume.

Question 34.

Give suitable explanation for the following facts about gases.

(a) Gases don’t settle at the bottom of a container

(b) Gases diffuse through all the space available to them and

(c) Explain with an increase in temperature

Answer:

(a) Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.

(b) When a sample of a gas introduced to one part of a closed container, its molecules very quickly disperse throughout the container, this process by which molecules disperse in space in response to differences in concentration is called diffusion. For e.g., you can smell perfume in a room, because it difluses into the air totally inside the room.

(c) Diffusion is faster at higher temperature because the gas molecules have greater kinetic energy. Since heat increase the motion, then diffusion happens faster.

Question 35.

Suggest why there ¡s no hydrogen (H_{2}) in our atmosphere. Why does the moon have no atmosphere?

Answer:

Hydrogen has a tendency to combine with oxygen from water vapour. Hence, the presence of hydrogen is negligible in the atmosphere. The gravitational pull in the moon is very less and hence, there is no atmosphere in the moon.

Question 36.

Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if –

(a) it is compressed to a smaller volume at constant temperature

(b) the temperature is raised while keeping the volume constant

(c) more gas is introduced into the same volume and at the same temperature

Answer:

(a) it a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) if more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. if the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 37.

Which of the following gases would you expect to deviate from ¡deal behaviour under conditions of low-temperature FCl_{2}, or Br_{2}? Explain.

Answer:

Bromine has a greater tendency to deviate from ideal behavior at low temperatures. The compressibility factor tends to deviate from unity for bromine.

Question 38.

Distinguish between diffusion and effusion.

Answer:

Diffusion:

- Diffusion is the spreading of molecules of a substance throughout a space or a second substance.
- Diffusion refers to the ability of the gases to mix with each other.
- E.g.. Spreading of something such as brown tea liquid spreading through the water in a tea cup.

Effusion:

- Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
- Effusion is a ability of a gas to travel through a small pin-hole.
- E.g., pouring out something like the soap studs bubbling out from a bucket of water.

Question 39.

Aerosol cans carry clear warning of heating of the can. Why?

Answer:

Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions. So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire or leave it in the direct sunlight. even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

- The gas pressure increases.
- More of the liquefied propellant turns into a gas.

Question 40.

When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?

Answer:

Helium floats because it is buoyant; its molecules are lighter than the nitrogen and oxygen molecules of our atmosphere and so they rise above it. In the car, it’s the air molecules that are actually getting pulled and pushed around by gravity as the result of the accelerating frame.

That means it moves in a direction opposite to the force on the surrounding air. Normally, the air is pulled downwards due to gravity, which pushes the balloon upwards. In this case, the surrounding air in the car is pulled forward by the deceleration of the car, which pushes the helium balloon backward.

Question 41.

Would it be easier to drink water with a straw on the top of Mount Everest?

Answer:

It is difficult to drink water with a straw on the top of Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water into the straw at the top of the mountain because gravity falls off gradually with height. The air pressure falls off, there isn’t enough atmospheric pressure to push the water up in the straw all the way to the mouth.

Question 42.

Write the Van der Waals equation for a real gas. Explain the correction term for pressure and volume.

Answer:

Van der Waals equation of state for real gases is

\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT

Correction term for pressure:

\(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\) is the pressure correction. It represents the intermolecular interaction that causes the non ideal behaviour.

Correction term for Volume:

V – nb is the volume correction. it is the effective volume occupied by real gas.

Question 43.

Derive the values of critical constants from the Van der Waals constants.

Answer:

Derivation of critical constants from the Van der Waals constants:

Van der Waals equation is,

\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT for 1 mole

From this equation, the values of critical constant P_{C}V_{C} and T_{C} arc derived in terms of a and b the Van der Waals constants.

\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V – b)\) = RT ………..(1)

On expanding the equestion (1)

P V + \(\frac {a}{V}\) – pb – \(\frac{\mathrm{ab}}{\mathrm{V}^{2}}\) – RT = 0 ………(2)

Multiplying eqestion (2) by \(\frac{V^{2}}{P}\),

equation (3) is rearranged in the powers of V

V^{3} – \(\left[\frac{\mathrm{RT}}{\mathrm{P}}+\mathrm{b}\right]\) V^{2} + \(\frac {aV}{P}\) –

= 0 ………..(4)

The above equation (4) is an cubic equation of V, which can have three roots. At the critical point. all the three values of V are equal to the critical volume V_{C}.

i.e. V = V_{C}.

V – V_{C} = O ……….(5)

(V – V_{C})^{3} = O ………(6)

(V^{3} – 3V_{C}V^{2} + 3V_{C}^{3}V – V_{C}^{3} = 0 ………(7)

As the equation (4) is identical with equation (7), comparing the ‘V’ ternis in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

substituting the values of Vc and Pc in equation (9)

Critical constant a and b can be calculated using Van der Waals Constant as follows:

Question 44.

Why do astronauts have to wear protective suits when they are on the surface of moon?

Answer:

Astronauts must wear spacesuits whenever they leave a spacecraft and are exposed to the environment of the moon. On the moon, there is no air to breathe and no air pressure. Moon is extremely cold and filled with dangerous radiation. Without protection, an astronaut would quickly die in space. Spacesuits are specially designed to protect astronauts from the cold, radiation, and low pressure in space. They also provide air to breathe. Wearing a spacesuit allows an astronaut to survive and work on the moon.

Question 45.

When ammonia combines with HCl, NH_{4} Cl is formed as white dense fumes. Why do more funies appear near HCl?

Answer:

- When ammonia combines with HCl, NH
_{4}Cl is formed as white dense fumes. The reaction takes place in neutralization between a weak base and a strong acid. - The property of the gas is diffusion.
- Diffusion of gases Ammonia and hydrogen chloride. Concentrated ammonia solution is placed on a pad in one end of a tube and concentrated HCl on the pad at the other.
- After about a minute, the gases diffuses far enough to meet and a ring of solid ammonium chloride is formed near the HCl end.

Question 46.

A sample of gas at 15°C at 1 atm has a volume of 2.58 dm^{3}. Vhen the temperature is raised to 38°C at I atm does the volume of the gas increase? if so, calculate the final volume.

Answer:

V_{2} = 2.78 dm^{3} i.e. volume increased from 2.58 dm^{3} to 2.78 dm^{3}.

Question 47.

A sample of gas has a volume of 8.5 dm^{3} at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 6.37 dm^{3}. What ¡s its initial temperature?

Answer:

T_{1} = 364.28 K

Question 48.

Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume of 37.6 dm^{3} at 298K, and the sample B is in a vessel of volume 16.5 dm^{3} at 298 K. Calculate the number of moles in sample B.

Answer:

n_{A} = 1.5 mol n_{B} = ?

V_{A} = 37.6 dm^{3} V_{B} = 16.5 dm^{3}

(T = 298 K constant)

Question 49.

Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 dm^{3} at 69.5°C, assuming ¡deal gas behaviour.

Answer:

n = 1.82 mole

V = 5.43 dm^{3}

T = 69.5 + 273 = 342.5

P = ?

PV = nRT

P = \(\frac {nRT}{V}\)

P = 94.25 atm.

Question 50.

Argon is an inert gas used in light bulbs to retard the vapourlzation of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure in atm.

Answer:

P_{1} = 1.2 atm

T_{1} = 18°C + 273 = 291 K

T_{2} = 85°C + 273 = 358 K

P_{2} = ?

\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 51.

A small bubble rises from the bottom of a lake, where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure is 1 atm. Calculate the final volume in (mL) of the bubble, if its initial volume is 1.5 mL.

Answer:

T_{1} = 6°C + 273 = 279 K

P_{1} = 4 atm V_{1} = 1.5m

T_{2} = 25°C + 273 = 298 K

P_{2} = 1 atm V _{1} = ?

\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V_{2} = 6.41 mol.

Question 52.

Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 x 10^{-3} dm^{3} of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?

Answer:

V = 154.4 x 10^{-3}dm^{3}

P = 742 mm of Hg

T = 298 K m = ?

n = \(\frac{PV}{RT}\) =

= 0.006 mol

n = \(\frac{PV}{RT}\)

n = \(\frac{Mass}{Molar Mass}\)

Mass = n x Molar mass

= 0.006 x 2.016

= 0.0121 g = 12.1 mg.

Question 53.

It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?

Answer:

Question 54.

A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure in the tank ¡s 9.21 atm. Calculate the partial pressure (in atm.) of each gas in the mixture.

Answer:

m_{O2} = 52.5 g

P_{O2} = ?

m_{CO2} = 65.1 g

P_{CO2} = ?

T = 300 K P = 9.21 atm

P_{O2} = X_{O2} x total pressure

P_{O2} = X_{O2} x Total pressure

= 0.53 x 9.21 atm = 4.88 atm

P_{CO2} = X_{CO2} x Total pressure

= 0.47 x 9.21 atm = 4.33 atm

Question 55.

A combustible gas Is stored in a metal tank at a pressure of 2.98 atm at 25 °C. The tank can withstand a maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal = 1100 K).

Answer:

T_{1} = 298 K;

P_{1} = 2.98 atm;

T_{2} = 1100K;

P_{2} = ?

\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P_{2} = \(\frac{P_{1}}{T_{1}} \times T_{2}\)

= \(\frac{2.98 \mathrm{atm}}{298 \mathrm{K}} \times 1100 \mathrm{K}\) = 11 atm

At 1100 K, the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

In Text Questions – Evaluate Yourself

Question 1.

Freon-I 2, the compound widely used in the refrigerator system as coolant causes depletion of ozone layer. Now It has been replaced by eco-friendly compounds. Consider 1.5 dm^{3} sample of gaseous freon at a pressure of 0.3 atm. If the pressure is changed to 1.2 atm. at a constant temperature, what will be the volume of the gas increased or decreased?

Answer:

Volume of freon (V_{1}) = 1.5 dm^{3}

Pressure (P_{1}) = 0.3 atm

‘T’ is constant

P_{2} = 1.2 atm

V_{2} = ?

P_{1}V_{1} = P_{2}V_{2}

V_{2} = \(\frac{P_{1} V_{1}}{P_{2}}\)

= 0.375 dm^{3}

Volume decreased from 1.5 dm^{3} to 0.375 dm^{3}

Question 2.

Inside a certain automobile engine, the volume of air in a cylinder is 0.375 dm^{3}, when the pressure is 1.05 atm. When the gas is compressed to a volume of 0.125 dm^{3} at the same temperature, what is the pressure of the compressed air?

Answer:

V_{1} = 0.375dm^{3}

V_{2} = 0.125 dm^{3}

P_{1} = 1.05 atm

P_{2} = ?

T – Constant

P_{1}V_{1} = P_{2}V_{2}

P_{2} = \(\frac{P_{1} V_{1}}{V_{2}}=\frac{10.5 \times 0.375}{0.125}\)

= 3.15 atm.

Question 3.

A sample of gas has a volume of 3.8 dm^{3} at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 2.27 dm^{3}. What is its initial temperature?

Answer:

V_{1} = 3.8 dm^{3}

T_{2} = 0°C = 273K

T_{1} = ? V_{2} = 2.27dm^{3}

\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

T_{1} = 457 K

Question 4.

An athlete in a kinesiology research study has his lung volume of 7.05 dm^{3} during a deep inhalation. At this volume the lungs contain 0.312 mole of air. During exhalation the volume of his Jung decreases to 2.35 dm^{3} How many moles of air does the athlete exhale during exhalation? (assume pressure and temperature remain constant)

Answer:

V_{1} = 7.05 dm^{3}

V_{2} = 2.35 dm^{3}

n_{1} = 0.312 mol

n_{1} =?

‘P’ and ‘T’ are constant

\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\)

n_{2} = 0.104 mol

Number of moles exhaled = 0.312 – 0.104 = 0.208 moles

Question 5.

A small bubble rises from the bottom of a lake, where the temperature and pressure are 8° C and 6.4 atm. to the water surface, where the temperature ¡s 25°C and pressure is 1 atm. Calculate the final volume in (ml) of the bubble, if its initial volume is 2.1 ml.

Answer:

T_{1} = 8°C = 8 + 273 = 281K

P_{1} = 6.4atm V_{1} = 2.1 mol

T_{2} = 25°C = 25 + 273 = 298 K

P_{2} = 1 atm

V_{2} = ?

\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V_{2} = 14.25 ml

Question 6.

(a) A mixture of He and O_{2} were used ¡n the ‘air’ tanks of underwater divers for deep dives. For a particular dive 12 dm^{3} of O_{2} at 298 K, I atm. and 46 dm3 of He, at 298 K, 1 aim. were both pumped into a 5 dm^{3} tank. Calculate the partial pressure of each gas and the total pressure In the tank at 298 K

Answer:

P = 1 atm

V_{total} = 5 dm^{3}

P_{O2} = X_{O2} x P_{total}

P_{He} = 3.54 atm

Question 6.

(b) A sample of solid KClO_{3} (potassium chlorate) was heated in a test tube to obtain O_{2} according to the reaction 2KClO_{3} → 2KCl_{(s)} + 3O_{2} The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mn of Hg at 300K. What is the partial pressure of the oxygen gas?

Answer:

2KCl_{3(s)} → 2KCl_{(s)} 3O_{3(g)}

P_{total} = 772 mm Hg

P_{H2O} = 26.7 mm Hg

P_{total} = P_{O2} + P_{H2O}

P_{O2} = P_{total} – P_{H2O}

P_{1} = 26.7 mm Hg

T_{1} = 300 K

T_{2} = 295 K

P_{2} = ?

\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P_{2} = 26.26 mm Hg

∴ P_{O2} = 772 – 26.26

= 745.74 mm Hg

Question 7.

A flammable hydrocarbon gas of particular volume is found to diffuse through a small hole in 1.5 minutes. Under the same conditions of temperature and pressure an equal volume of bromine vapour takes 4.73 min to diffuse through the same hole. Calculate the molar mass of the unknown gas and suggest what this gas might be, (Given that molar mass of bromine = 159.8 g/ mole)

Answer:

t_{1} = 1.5 minutes (gas)_{hydrocarbon}

t_{2} = 4.73 minutes (gas)_{Bromi}

n (12) + (2n + 2) 1 = 16 (general formula for hydrocarbon C_{n}H_{2n+2})

12n + 2n + 2 = 16

14n = 16 – 2

14n = 14

n = 1

The hydrocarbon is C_{1}H_{2(1) + 2} = CH_{4}

Question 8.

Critical temperature of H_{2}O, NH_{3} and CO_{2} are 647.4, 405.5 and 304.2 K, respectively. When we start cooling from a temperature of 700 K which will liquefy first and which will liquefy finally?

Answer:

Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.

∴ When cooling starts from 700 K, H_{2}O vill liquefied first, then followed by ammonia and finally carbon dioxide will liquefied.

In-Text Example Problems

Question 9.

In the below figure, let us find the missing parameters [volume in (b) and pressure in (c)]

P_{1} = 1 atm P_{2} = 2 atm P_{3} = ? atm

V_{1} I dm3 V_{2} =? dm3 V_{3} = 0.25 dm^{3}

T_{1} = 298 K T_{2} = 298 K T_{3} = 298 K

Solution:

According to Boyle’s law, at constant temperature for a given mass of gas at constant temperature,

P_{1}V_{1} = P_{2}V_{2} = P_{3}V_{3}

I atm x 1 dm^{3} = 2 atm x V_{2} = P_{3}x 0.25 dm^{3}

∴ 2 atm x V_{2} = 1 atm x 1 dm^{3}

V_{2} = 0.5 dm^{3}

and P_{3} x 0.25 dm^{3} = 1 atm x 1 dm^{3}

P_{3} = 4atm

Question 10.

In the below figure, let us find the missing parameters [volume in (b) and temperature in (c)]

P_{1} = 1 atm P_{2} = 1 atm P_{3} = 1 atm

V_{1} = 0.3dm^{3} V_{2} = ?dm^{3} V_{3} = 0.15dm^{3}

T_{1} = 200K T_{2} = 300 K T_{3} = ? K

Answer:

According to Charles law,

T_{3} = 100k

Question 11.

Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm^{3} at 70°C assuming It is an ideal gas.

Answer:

We will use the ideal gas equation for this calculation as below:

P = \(\frac {nRT}{V}\)

= 9.39 atm.

Question 12.

A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm. at a fixed temperature. Solve this problem using Dalton’s law.

Answer:

P_{Ne} = X_{Ne} P_{Total}

P_{Ne} = X_{Ne} P_{Total} = 0.595 x 2 = 1.19 atm.

P_{Ne} = X_{Ne} P_{Total} = 0.093 x 2 = 0. 186 atm.

P_{Ne} = X_{Ne} P_{Total} = 0.312 x 2 = 0.624 atm.

Question 13.

An unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperature and pressure. Calculate the molar mass of the unknown gas.

Answer:

Question 14.

If a scuba diver takes a breath at the surface filling his lungs with 5.82 dm3 of air what volume will the air in his lungs occupy when he drives to a depth, where the pressure ¡s 1.92 atm. (assume temperature is constant and the pressure at the surface is exactly 1 atm.)

Solution :

Temperature = Constant

Pressure at the surface = 1 atm – P_{1}

Pressure at the depth = 1.92 atm – P_{2}

Vdlume of air breathing at the surface of the air = 5.82 dm^{3} – V_{1}

Volume of air breathing at the depth = V_{2} = ?

V_{1} = 3.03 dm^{3}

The volume of air scuba diver’s lung occupy = 3.03 dm^{3}

Question 15.

Inside a certain automobile engine, the volume of air in a cylinder is 0.475 dm^{3}, when the pressure is 1.05 atm. When the gas is compressed, the pressure increased to 5.65 atm. at the same temperature. What is the volume of compressed air?

Solution:

Volume of air in the cylinder 0.475 dm^{3} – V_{1}

Pressure of air P_{1} = 1.05 atm

Increased pressure P_{2} = 5.65 atm

Volume of air compressed V_{2} = ?

P_{1}V_{1} = P_{2}V_{2}

V_{2} = 0.08827 dm^{3}

Compressed volume of air = 0.08 827 dm^{3}

### Samacheer Kalvi 11th Chemistry Gaseous State Additional Questions Solved

I. Choose the correct answer.

Question 1.

The approximate volume percentages of nitrogen and oxygen in the atmosphere of air are _______ and ______ respectively.

(a) 21, 68

(b) 21, 78

(c) 78, 2

(d) 80, 21

Answer:

(c) 78, 2

Question 2.

The average kinetic energy of the gas molecule is …………

(a) inversely proportional to its absolute temperature

(b) directly proportional to its absolute temperature

(c) equal to the square of its absolute temperature

(d) All of the above

Answer:

(b) directly proportional to Its absolute temperature

Question 3.

The compound widely used in the refrigerator as coolant is

(a) Freon-2

(b) Freon-12

(c) Freon-13

(d) Freon-14

Answer:

(b) Freon-12

Question 4.

At constant temperature, the pressure of the gas is reduced to one-third, the volume

(a) reduce to one-third

(b) increases by three times

(c) remaining the same

(d) cannot be predicted

Answer:

(b) increases by three times

Question 5.

Hydrogen is placed in the ______ of the periodic table.

(a) Group -1

(b) group-17

(c) group-18

(d) group-2

Answer:

(a) Group -1

Question 6.

Viscosity of a liquid is a measure of ……………

(a) repulsive forces between the liquid molecules

(b) frictional resistance

(c) intermolecular forces between the molecules

(d) none of the above

Answer:

(b) frictional resistance

Question 7.

At constant temperature for a given mass, for each degree rise in temperature, all gases expand by of their volume at 0°C.

(a) 273

(b) 298

(c) \(\frac{1}{273}\)

(d) \(\frac{1}{298}\)

Answer:

(c) \(\frac{1}{273}\)

Question 8.

In Vander Waals equation of state for a non-ideal gas the net force of attraction among the molecules is given by ………..

(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

(b) P + \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

(c) P – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

(d) – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

Answer:

(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

Question 9.

‘At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature”. This statement is

(a) Charle’s law

(b) Boyle’s law

(c) Gay-Lussac’s law

(d) Dalton’s law

Answer:

(c) Gay-Lussac’s law

Question 10.

Which of the following gases will have the lowest rate of diffusion?

(a) H_{2}

(b) N_{2}

(c) F_{2}

(d) O_{2}

Answer:

(c) F_{2}

Question 11.

A mixture of gases containing 4 moles of hydrogen and 6 moles of oxygen. The partial pressure of hydrogen, if the total pressure is 5 atm, is

(a) 10 atm

(b) 0.4 atm

(c) 20 atm

(d) 2 atm

Answer:

(d) 2 atm

Question 12.

Which of the following is a diatomic gas in nature?

(a) Oxygen

(b) Ozone

(c) Helium

(d) Radon

Answer:

(a) Oxygen

Question 13.

The property of a gas which involves the movement of the gas molecules through another gas is called

(a) effusion

(b) dissolution

(c) diffusion

(d) expansion

Answer:

(c) diffusion

Question 14.

Among the following groups which contains monoatomic gases?

(a) Group 17

(b) Group 18

(c) Group 1

(d) Group 15

Answer:

(b) Group 18

Question 15.

The rate of diffusion of a gas is inversely proportional to the

(a) square of molar mass

(b) square root of density

(c) square root of molar mass

(d) square of density

Answer:

(c) square root of molar mass

Question 16.

Which of the following gas is essential for our survival?

(a) N_{2}

(b) H_{2}

(c) O_{2}

(d) He

Answer:

(c) O_{2}

Question 17.

The deviation of real gases from ideal behavior is measured in terms of

(a) expansivity factor

(b) molar mass

(c) pressure

(d) compressibility factor

Answer:

(d) compressibility factor

Question 18.

Which of the following is not chemically inert?

(a) Helium

(b) Oxygen

(c) Argon

(d) Krypton

Answer:

(b) Oxygen

Question 19.

Match the List-I and List-II using the correct code given below the list.

Answer:

Question 20.

Ideal gas equation for ‘n’ moles is

(a) \(\frac{P}{R}=\frac{n T}{V}\)

(b) PV = \(\frac{n R}{T}\)

(c) \(\frac{V}{T}=\frac{n P}{R}\)

(d) \(\frac{P}{T}=\frac{n R}{V}\)

Answer:

(d) \(\frac{P}{T}=\frac{n R}{V}\)

Question 21.

The SI unit of pressure is ………..

(a) Nm^{-2} Kg^{-1}

(b) Pascal

(c) bar

(d) atmosphere

Answer:

(b) Pascal

Question 22.

The volume correction introduced by Van der Waals is, V_{ideal} =

(a) V – nb

(b) V + nb

(c) \(\frac{V}{n b}\)

(d) b – nV

Answer:

(b) V + nb

Question 23.

The instrument used for measuring the atmospheric pressure is …………….

(a) lactometer

(b) barometer

(c) electrometer

(d) ammeter

Answer:

(b) barometer

Question 24.

The unit of Van der Waals constant V is

(a) atm lit mol^{-1}

(b) atm lit^{2} mol^{-2}

(c) atm lit^{-2} mol^{2}

(d) atm lit^{-1} mol^{2}

Answer:

(b) atm lit^{2} mol^{-2}

Question 25.

Mathematical expression of Boyle’s law is ………….

(a) P_{1}V_{1} = P_{2}V_{2}

(b) \(\frac {P}{V}\) = Constant

(c) \(\frac {V}{T}\) = Constant

(J) \(\frac {P}{T}\) = Constant

Answer:

(a) P_{1}V_{1} = P_{2}V_{2}

Question 26.

Statement-I: If the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.

Statement-II: If the volume is halved, the density of the gas is doubled.

(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I

(c) Statement-I is correct but Statement-II is wrong

(d) Statement-I is wrong but Statement-II is correct

Answer:

(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 27.

The gas used in the pressure-volume isotherm study of Andrew’s experiment is

(a) N_{2}

(b) H_{2}S

(c) NH_{3}

(d) CO_{2}

Answer:

(d) CO_{2}

Question 28.

Which one of the following is absolute zero?

(a) 293 K

(b) 273 K

(c) – 273.15°C

(d) 0°C

Answer:

(c) – 273.15°C

Question 29.

The relationship between critical volume and Vander waals constant is

(a) V_{c} =\(\frac{a}{R b}\)

(b) V_{c} = 3b

(c) V_{c} = \(\frac{8 a}{27 R b}\)

(d) V_{c} = 8b

Answer:

(b) V_{c} = 3b

Question 30.

The ideal gas equation is …………..

(a) PV = RT for 1 mole

(b) P_{1} V_{1} = P_{2}V_{2}

(c) \(\frac {P}{T}\) = R

(d) P = P_{2} + P_{2} + P_{2}

Answer:

(a) PV = RT for 1 mole

Question 31.

In the equation PV = nRT, which one cannot be numerically equal to R?

(a) 8.31 × 10^{7} ergs K^{-1} mol^{-1}

(b) 8.31 × 10^{7} dynes cm K^{-1} mol^{-1}

(c) 8.317J K^{-1} mol^{-1}

(d) 8.317L atm K^{-1} mol^{-1}

Answer:

(d) 8.317L atm K^{-1} mol^{-1}

Question 32.

Mathematical expression of Graham’ s law is ……………….

Answer:

Question 33.

The number of moles of H_{2} in 0.224 L of hydrogen gas at STP (273 K, 1 atm) assuming ideal gas behavior is

(a) 1

(b) 0.1

(c) 0.01

(d) 0.001

Answer:

(c) 0.01

Question 34.

The value of compression factor Z is equal to ………….

(a) \(\frac {nRT}{PV}\)

(b) \(\frac {PV}{RT}\)

(c) PV x nRT

(d) \(\frac {PV}{nRT}\)

Answer:

(d) \(\frac {PV}{nRT}\)

Question 35.

To what temperature must a neon gas sample be heated to double its pressure, if the initial volume of gas at 75°C is decreased by 15.0% by cooling the gas?

(a) 319°C

(b) 592°C

(c) 128°C

(d) 60°C

Answer:

(a) 319°C

Question 36.

The value of critical temperature of carbon dioxide is …………

(a) 273 K

(b) 303.98 K

(c) 373 K

(d) – 80°C

Answer:

(b) 303.98 K

Question 37.

Match the List-I and List-II using the correct code given below the list.

Answer:

Question 38.

The rate of diffusion of gases A and B of molecular weight 36 and 64 are in the ratio

(a) 9 : 16

(b) 4 : 3

(c) 3 : 4

(d) 16 : 9

Answer:

(b) 4 : 3

Question 39.

The temperature below which a gas obey Joule Thomson effect is called …………..

(a) critical temperature

(b) standard temperature

(c) inversion temperature

(d) normal temperature

Answer:

(c) Inversion temperature

Question 40.

Which of the following is true about the gaseous state?

(a) Thermal energy = molecular interaction

(b) Thermal energy >> molecular interaction

(c) Thermal energy << molecular interaction

(d) molecular forces >> those in liquids

Answer:

(b) Thermal energy >> molecular interaction

Question 41.

The temperature produced in adiabatic process of liquefaction is …………..

(a) zero kelvin

(b) -273 K

(c) 10^{-4} K

(d) 10^{4} K

Answer:

(c) 10^{-4} K

Question 42.

Gas deviates from ideal gas nature because molecules

(a) are colourless

(b) attract each other

(c) contain covalent bond

(d) show Brownian movement

Answer:

(b) attract each other

Question 43.

The compressibility factor for an ideal gas is …………….

(a) 1.5

(b) 2

(c) 1

(d) x

Answer:

(c) 1

Question 44.

Which of the following pair will diffuse at the same rate?

(a) CO_{2} and N_{2}O

(b) CO_{2} and NO

(c) CO_{2} and CO

(d) N_{2}O and NO

Answer:

(a) CO_{2} and N_{2}O

Question 45.

What are the most favourable conditions to liquefy a gas?

(a) High temperature and low pressure

(b) Low temperature and high pressure

(c) High temperature and high pressure

(d) Low temperature and low pressure

Answer:

(b) Low temperature and high pressure

Question 46.

A person living in Shimla observed that cooking food with using pressure cooker takes more time. The reason for this observation is that at high altitude …………..

(a) pressure increases

(b) temperature decreases

(c) pressure decreases

(a) temperature decreases

Answer:

(c) pressure decreases

Question 47.

Statement-I: At constant temperature PV vs V plot for real gases is not a straight line.

Statement-II : At high pressure, all gases have Z >1, but at intermediate pressure most gases have Z <1.

(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I

(c) Statement-I is correct but Statement-Il is wrong

(d) Statement-I is wrong but Statement-Il is correct .

Answer:

(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 48.

Statement-I: Gases do not liquefy above their critical temperature, even on applying high press ure.

Statement-II: Above critical temperature, the molecular speed is high and intermolecular

attractions cannot hold the molecules together because they escape because of high speed.

(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I

(c) Statement-I is correct but Statement-II is wrong

(d) Statement-I is wrong but Statement-II is correct

Answer:

(c) Statement-I and II are correct and Statement-IIs the correct explanation of Statement-I

Question 49.

NH_{3} gas is liquefied more easily than N_{2}. Hence

(a) Van der Waals constant a and b of NH_{3} > that of N_{2}

(b) van der Waals constants a and b of NH_{3} < that of N_{2}

(c) a(NH_{3}) > a(N_{2}) but b(NH3) < b(N_{2})

(d) a(NH_{3}) < a(N_{2}) but b(NH_{3}) > b(N_{2})

Answer:

(c) a(NH_{3}) > a(N_{2}) but b(NH3) < b(N_{2})

Question 50.

In a closed flask of 5 liters, 1.0 g of H_{2} is heated from 300 to 600 K, which statement is not correct’?

(a) pressure of the gas increases

(b) the rate of the collusion increase

(c) the number of moles of gas increases

(d) the energy of gaseous molecules increases

Answer:

(c) the number of moles of gas increases

Question 51.

Match the List-I and List-II using the correct code given below the list.

Answer:

Question 52.

Consider the following statements.

(i) All the gases have higher densities than liquids and solids.

(ii) All gases occupy zero volume at absolute zero.

(iii) At very low pressure all gases exhibit ideal behaviour.

Which of the above statement is/are not correct?

(a) (i) only

(b) (ii) only

(c) (iii) only

(d) (ii) and (iii) only

Answer:

(a) (i) only

Question 53.

Which of the following gas is present maximum in atmospheric air?

(a) O_{2}

(b) N_{2}

(c) H_{2}

(d) radon

Answer:

(b) N_{2}

Question 54.

Which law is used in the process of enriching the isotope of U^{235} from other isotopes?

(a) Boyle’s law

(b) Dalton’s law of partial pressure

(c) Graham’s law of diffusion

(d) Charles’ law

Answer:

(c) Graham’s law of diffusion

**Samacheer Kalvi 11th Chemistry Gaseous State 2 – Mark Questions**

Question 1.

Define Pressure? Give its unit?

Answer:

Pressure is defined as force divided by the area to which the force is applied. The SI unit of pressure is pascal which is defined as 1 Newton per square meter (Nm^{-2}).

Pressure = \(\frac{\text { Force }\left(\mathrm{N}(\text { or }) \mathrm{Kg} \mathrm{ms}^{-2}\right)}{\text {Area }\left(\mathrm{m}^{2}\right)}\)

Question 2.

Distinguish between a gas and a vapour.

Answer:

- Gas: A substance that is normally in a gaseous state at ordinary temperature and pressure. e.g., Hydrogen.
- Vapour: The gaseous form of any substance that is a liquid or solid at normal temperature and pressure. e.g., At 298 K and 1 atm, water exist as water vapour.

Question 3.

State Dalton’s law of partial pressure.

Answer:

The total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton s law of partial pressures.

For a mixture containing three gases 1, 2, and 3 with partial pressures p_{1}, p_{2,} and p_{3} in a container with volume V, the total pressure Ptotal will be given by

P_{total} = p_{1} + p_{2} + p_{3}

Question 4.

Define atmospheric pressure. What is its value?

Answer:

- The pressure exerted on a unit area of Earth by the column of air above it is called atmospheric pressure.
- The standard atmospheric pressure = 1 atm.
- 1 atm = 760 mm Hg.

Question 5.

Define the Critical temperature (T_{c}) of a gas?

Answer:

Critical temperature (T_{c}) of a gas is defined as the temperature above which it cannot be liquefied even at high pressure.

Question 6.

Most aeroplanes cabins are artificially pressurized. Why?

Answer:

The pressure decreases with the increase in altitude because there are fewer molecules per unit volume of air. Above 9200 m (30,000 ft), for example, where most commercial aeroplanes fly, the pressure is so low that one could pass out for lack of oxygen. For this reason most aeroplanes cabins arc artificially pressurized.

Question 7.

How do you understand the PV relationship?

Answer:

The PV relationship can be understood as follows. The pressure is due to the force of the gas particles on the walls of the container. If a given amount of gas is compressed to half of its volume, the density is doubled and the number of particles hitting the unit area of the container will be doubled. Hence, the pressure would increase twofold.

Question 8.

State Charles’ law.

Answer:

Charles’ law:

For a fixed mass of a gas constant pressure, the volume is directly proportional to temperature (K).

Mathematically V – T at constant P and n. (or) \(\frac {V}{T}\) = Constant (or) \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) = Constant

Question 9.

A gas cylinder can withstand a pressure of 15 aim. The pressure of the cylinder is measured 12 atm at 27°C. Upto which temperature limit the cylinder will not burst?

Answer:

Cylinder will burst at that temperature when it attains the pressure of 15 atm

P_{1} = 12 atm

T = 27° C = (27 + 273)K = 300K

P_{2} = 15 atm; T = ?

\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

T_{2} = \(\frac{15 \times 300}{12}\) = 375 K

= (375 – 273)°C = 102°C

Question 10.

State Avogadro’s hypothesis.

Answer:

Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. Mathematically V ∝ n

\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant

Question 11.

State Charles law.

Answer:

For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature.

\(\frac{V}{T}\) = constant (at constant pressure)

Question 12.

What are the applications of Dalton’s law of partial pressure?

Answer:

1. Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab, the values are reported as partial pressures.

Gas – Normal range

P_{CO2} 35 – 45mm of Hg

P_{O2} 80 – 100 mm of Hg

2. When gas is collected by downward displacement of water, the pressure of dry vapour collected is computed using Dalton’s law

Question 13.

How can you identify a heavy smoker with the help of Dalton’s law?

Answer:

Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab. The values are reported as partial pressures.

A heavy smoker may be expected to have low O_{2} and huge CO_{2} partial pressures.

Question 14.

State Gay-Lussac law.

Answer:

At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.

P ∝ T (at constant volume)

Question 15.

Helium diffuses more than air. Give reason.

Answer:

Take two balloons, one is filled with air and another with helium. After one day, the helium balloon was shrunk, because helium being lighter diffuses out faster than the air

Question 16.

Distinguish between diffusion and effusion.

Answer:

The property of gas that involves the movement of the gas molecules through another gas is called diffusion. Effusion is another process in which a gas escapes from a container through a very small hole.

Question 17.

What is the compression factor?

Answer:

The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT.

This is termed as a compression factor.

Compression factor = Z = \(\frac {PV}{nRT}\)

For ideal gases Z = 1 at all temperatures and pressures.

Question 18.

- Define critical temperature.
- What is the critical temperature of CO
_{2}gas?

Answer:

- The temperature below which a gas can be liquefied by the application of pressure is known as critical temperature.
- The critical temperature of CO
_{2}gas is 303.98 K.

Question 19.

- Define critical pressure.
- What is the critical pressure of CO
_{2}gas?

Answer:

- Critical temperature (P
_{C}) of a gas is defined as the minimum pressure required to liquefy. - The critical pressure of CO
_{2}is 73 atm.

Question 20.

CO_{2} gas cannot be liquefied at room temperature. Give reason.

Answer:

Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO_{2} has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 3O_{3} K. At room temperature, (critical temperature) even by applying large amount of pressure CO_{2} cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO_{2} remains as gas.

Question 21.

What is meant by the Joule-Thomson effect?

Answer:

The phenomenon of lowering of temperature when a gas is made to expand adiabatically form a region of high pressure into a region of low pressure is known as Joule-Thomson effect.

Question 22.

Define inversion temperature.

Answer:

The temperature below which a gas obey Joule-Thomson effect is called inversion temperature (T_{i}).

Ti = \(\frac {2a}{Rb}\)

Question 23.

State and explain Boyle’s law. Represent the law graphically.

Answer:

It states that, the pressure of a fixed mass of a gas is inversely proportional to its volume if temperature is kept constant.

P – \(\frac {1}{V}\)

PV = Constant (n and T are constant)

P_{1}v_{1} = P_{2}V_{2}

Graphical Representation:

Question 24.

Give an expression for the van der Waals equation. Give the significance of the constants used in the equation. What are their units?

Answer:

\(\left(P+\frac{n^{2} a}{V^{2}}\right)\) (V- nb) = nRT

Where n is the number of moles present and ‘a’’ b’ are known as van der Waals constants.

Significance of Van der Waals constants:

Van der Waals constant ‘a’:

‘a’ is related to the magnitude of the attractive forces among the molecules of a particular gas. Greater the value of ’a’, more will be the attractive forces.

Unit of ’a’ = L^{2} mol^{-2}

Van der Waals constant ‘b’:

‘b’ determines the volume occupied by the gas molecules which depends upon size of molecule.

Unit of ‘b’ = L mol^{-1}

Question 25.

What are ideal and real gases? Out of CO_{2} and NH_{3} gases, which is expected to show more deviation from the ideal gas behaviour?

Answer:

Ideal gas:

A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.

Real gases:

Gases which deviate from ideal gas behaviour are known as real gases. NH_{3} is expected to show more deviation. Since NH_{3} is polar in nature and it can be liquefied easily.

**Samacheer Kalvi 11th Chemistry Gaseous State 3 – Mark Questions**

Question 1.

Explain the graphical representation of Boyle’s law.

Answer:

Boyle’s law states that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure. V ∝ (T and n are fixed).

If the pressure of the gas increases, the volume will decrease and if the pressure of the gas decreases, the volume will increase. So PV = Constant.

Question 2.

What are the different methods of liquefaction of gases?

Answer:

There are different methods used for the liquefaction of gases:

- In Linde’s method, the Joule-Thomson effect is used to get liquid air or any other gas.
- In Claude’s process, the gas is allowed to perform mechanical work in addition to the Joule-Thomson effect so that more cooling is produced.
- In the Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10
^{-4}K i.e., as low as 0 K can be achieved.

Question 3.

Explain Charles’ law with an experimental illustration.

Answer:

Charles’ law states that for a fixed mass of a gas at constant pressure, the volume is directly proportional to temperature (K).

V ∝ T (or) \(\frac {V}{T}\) = Constant

Volume vs Temperature:

If a balloon is moved from an ice water bath to a boiling water bath, the gas molecules inside move faster due to increased temperature and hence the volume increases.

Question 4.

Explain the graphical representation of Charles’ law.

Answer:

1. Variation of volume of the gas sample with temperature at constant pressure.

2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P_{1} to P_{5}.

3. i.e. P_{1}<P_{2} < P_{3} <P_{4} < P_{5} . When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.

4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.

5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 5.

Explain graphicl representation of Gay Lussac’s law

Answer:

Gay Lussac’s law

At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.

P – T (or) \(\frac {P}{T}\) = Constant

It can be graphically represented as shown here:

Lines in the pressure vs temperature graph are known as isochores (constant volume) of a gas.

Question 6.

Explain the graphical representation of Avogadro’s hypothesis.

Answer:

Avogadro’s hypothesis states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

V ∝ n

\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant.

Where V_{1} and are the volume and number 10- of moles of a gas and V_{2} and n_{2} are the different set of values of volume and number of moles of the same gas.

A better example to illustrate Avogadro’s Number of Moles (n) hypothesis is to observe the effect of pumping more gas into a balloon. When more gas molecules (particularly CO_{2}) are passed, the volume of the balloon increases. The pressure and temperature stay constant as the balloon inflates, so the increase in volume is due to the increase in the quantity of gas inside the balloon.

Question 7.

The vapour pressure of water at 80°C is 355.5 mm of Hg. A 100 mL vessel contains water saturated with O_{2} at 80°C, the total pressure being 760 mm of Hg. The contents of the vessel were pumped into a 50 mL vessel at the same temperature. What is the partial pressure of O_{2}?

Answer:

P_{total} = P_{H2O} + P_{O2}

P_{O2} = 760 – 355.5 = 404.5

When the contents were pumped into 50 mL vessel

P_{1} = 404.5

V_{1} = 100 mL

P_{2} =?

V_{2} = 50 mL

P_{1}V_{1} = P_{2}V_{2}

P_{2} = \(\frac{404.5 \times 100}{50}\) = 809 mm

Thus, PO_{2} in 50 mL vessel = 809 mm

Question 8.

Derive the various values of R, gas constant.

Answer:

1. For standard conditions in which P is 1 atm, volume 22.4 14 dm^{3} for 1 mole at 273.15 K.

= 0.08205 7 dm^{3} atm mol^{-1} K^{-1}

2. Where P = 10^{5} Pascal, V = 22.71 x 10^{-3} m^{3} for I mole of a gas at 273.15 K.

= 8.314 pa m^{3} K^{-1} mol^{-1}

=8.314x 10^{-2}bar dm^{3} K^{-1}mol^{-1}

R = 8.314 J K^{-1} mol^{-1}.

Question 9.

What is meant by Boyle’s temperature (or) Boyle’s point? How is it related to the compression point?

Answer:

(1) Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.

(2) The Boyle point varies with the nature of the gas.

(3) Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.

(4) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperatures, the real gases behave as ideal gases.

Question 10.

State and explain Dalton’s law of partial pressure.

Answer:

John Dalton stated that “the total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton’s law of partial pressures, i.e., for a mixture containing three gases 1, 2, and 3 with partial pressures p_{1}, p_{2,} and p_{3} in a container with volume V, the total pressure P_{total} will be given by P = p_{1} + p_{2} + p_{3}

**Samacheer Kalvi 11th Chemistry Gaseous State 5 – Mark Question**

Question 1.

How CH_{4} He and NH_{3} are deviating from ideal behaviour? (or) Explain how real gases deviate from ideal behaviour.

Answer:

1. The gases which obey gas equation PV = nRT are known as ideal gases. The gases which do not obey PV = nRT are known as real gases.

2. The gas laws and the kinetic theory are based on the assumption that molecules in the gas phase occupy negligible volume (assumption 1) and that they do not exert any force on one another either attractive or repulsive (assumption 2). Gases whose behaviour is consistent with these assumptions are said to exhibit ideal behaviour.

3. The following graph shows RT plotted against P for three real gases and an ideal gas at a given temperature.

4. According to ideal gas equation, PV/RT is equal to n. Plot PV/RT versus P for ‘n’ moles oigas at 0°C. For1 mole of an ideal gas PV/ RT is equal to 1 irrespective of the pressure of the gas.

5. For real gases, we observe various deviations from ideal behaviour at high pressure. At very low pressure, all gases exhibit ideal behaviour, ie. PV/RT values all converge to n as P approaches zero.

6. For real gases, this is true only at moderate low pressures. (≤ 5 atm) significant variation occurs as the pressure increases. Attractive forces operate among molecules at relatively short deviation.

7. At atmospheric pressure, the molecules in a gas are far apart and attractive forces arc negligible. At high pressure, the density of the gas increases and the molecules are much closer to one another. Intermolecular forces can be significant enough to affect the motion of the molecules and the gas will not behave ideally.

Question 2.

Derive Van der Waals equation of state.

Answer:

1. Consider the effect of intermolecular forces on the pressure exerted by a gas form the following explanation.

2. The speed of a molecule that is moving toward the wall of a container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is Q lower than the pressure the gas would exert, if it behave ideally.

Where ‘a’ is the proportionality constant and depends on the nature of the gas and n and V are the number of moles and volume of the container and respectively an^{2}/ V^{2} is the correction term.

3. The frequency of encounters increases with the square of the number of molecules per unit volume n^{2}/ V^{2}. Therefore an^{2}/ V^{2} represents the intermolecular interaction that causes non-ideal behaviour.

4. Another correction is concerned with the volume ¿ccupied by the gas molecules. ‘V’ represents the volume of the container. As every individual molecule of a real gas occupies certain volume, the effective volume V- nb which is the actually available for the gas, n is the number of moles and b is a constant of gas.

5. Hence Van der Waals equation of state for real gases are given as \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT Where a and b are Van der Waals constants.

Question 3.

Explain about Andrew’s experimental isotherms of CO_{2} gas.

Answer:

1. Andrew plotted isotherms of carbon dioxide at different temperatures. it is then proved that many real gases behave in a similar manner like CO_{2}.

2. At a temperature of 303.98 K, CO_{2} remains as a gas. Below this temperature, CO_{2}, turns into liquid CO_{2} at 7 3atm. It is called the critical temperature of CO_{2}.

3. At 303.98 K and 73 atm pressure, CO_{3},, becomes a liquid but remains a gas at higher temperature.

4. Below the critical temperature, the behaviour of CO_{2} is different. For example, consider an isotherm of CO_{2} at 294.5 K, it is a gas until the point B, is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist. At C, the gas is completely condensed.

5. If the pressure is higher than at C, only the liquid is compressed so, a steep rise in pressure is observed. Thus, there exist a continuity of state.

6. A gas below the critical temperature can be liquefied by applying pressures.

**Activity – 1**

The table below contains the values of pressure measured at different temperatures for 1 moie of an ideal gas. Plot the values in a graph and verify the Gay Lussac’s law. [Lines in the pressure vs temperature graph are known as iso chores (constant volume) of a gas]

Solution:

Gay Lussac’s law at constant volume = \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

If temperature increases, pressure also increases.

So \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)