Kinetic Theory of Gases and ldeal Gas | Physics JEE previous year questions with Solutions

Q 21 :

At 300 K, the rms speed of oxygen molecules is $\sqrt{\frac{α + 5}{α}}$ times to that of its average speed in the gas. Then, the value of $α$ will be (use $π = \frac{22}{7}$ ) [2023]

32
28
24
27

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(2)

$\sqrt{\frac{3 R T}{M}} = \sqrt{\frac{α + 5}{α}} \sqrt{\frac{8}{π} \frac{R T}{M}}$

$3 = \frac{α + 5}{α} \frac{8}{π} \Rightarrow α = 28$

Q 22 :

Heat energy of 735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but does not oscillate. The increase in the internal energy of the gas will be [2023]

525 J
441 J
572 J
735 J

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(2)

$Δ Q = n C_{p} Δ T = 735 J$

$\Rightarrow \frac{5 n R Δ T}{2} = 735 J$

$Δ U = n C_{v} Δ T = \frac{3}{2} (n R Δ T) = \frac{3}{2} \times \frac{2}{5} \times 735 = 441 J$

Q 23 :

The average kinetic energy of a molecule of the gas is [2023]

proportional to absolute temperature
proportional to volume
proportional to pressure
dependent on the nature of the gas

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(1)

Translational K.E. on average of a molecule is $\frac{3}{2} k T,$ which is independent of nature, pressure, and volume.

Q 24 :

The temperature of an ideal gas is increased from 200 K to 800 K. If the r.m.s. speed of the gas at 200 K is $v_{0}$ , then the r.m.s. speed of the gas at 800 K will be [2023]

$v_{0}$
$4 v_{0}$
$\frac{v_{0}}{4}$
$2 v_{0}$

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(4)

$V_{rms} = \sqrt{\frac{3 R T}{M}} \Rightarrow V_{rms} \propto \sqrt{T}$

Increasing the temperature 4 times, the rms speed gets doubled.

Q 25 :

An air bubble of volume ${1 cm}^{3}$ rises from the bottom of a lake 40 m deep to the surface at a temperature of $12 ° C$ . The atmospheric pressure is $1 \times 10^{5} Pa$ , the density of water is $1000 {kg/m}^{3}$ and $g = 10 {m/s}^{2}$ . There is no difference of the temperature of water at the depth of 40 m and on the surface. The volume of the air bubble when it reaches the surface will be [2023]

2 ${cm}^{3}$
4 ${cm}^{3}$
5 ${cm}^{3}$
3 ${cm}^{3}$

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(3)

$P = P_{0} + ρ g h = 10^{5} Pa + 10^{3} \times 10 \times 40 = 5 \times 10^{5} Pa$

At T is constant, $P V = P_{0} V_{0}$

$\Rightarrow 5 \times 10^{5} Pa \times 1 {cm}^{3} = 10^{5} Pa \times V_{0}$ $\Rightarrow V_{0} = 5 {cm}^{3}$

Q 26 :

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and the third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square speed ( $v_{rms}$ ) and choose the correct answer from the options given below: [2023]

$v_{rms} (mono) = v_{rms} (dia) = v_{rms} (poly)$
$v_{rms} (mono) > v_{rms} (dia) > v_{rms} (poly)$
$v_{rms} (dia) < v_{rms} (poly) < v_{rms} (mono)$
$v_{rms} (mono) < v_{rms} (dia) < v_{rms} (poly)$

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(2)

$v_{rms} (mono) = \sqrt{\frac{3 R T}{4 \times 10^{- 3}}}$

$v_{rms} (dia) = \sqrt{\frac{3 R T}{71 \times 10^{- 3}}}$

$v_{rms} (poly) = \sqrt{\frac{3 R T}{146 \times 10^{- 3}}}$

$So the correct relation is$

$v_{rms} (mono) > v_{rms} (dia) > v_{rms} (poly)$

Q 27 :

The root mean square speed of molecules of nitrogen gas at $27 ° C$ is approximately:

(Given mass of a nitrogen molecule $= 4.6 \times 10^{- 26} kg$ and take Boltzmann constant $k_{B} = 1.4 \times 10^{- 23} {JK}^{- 1}$ ) [2023]

523 m/s
1260 m/s
91 m/s
27.4 m/s

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(1)

$V_{rms} = \sqrt{\frac{3 k_{B} T}{m}} = \sqrt{\frac{3 \times 1.4 \times 10^{- 23} \times 300}{4.6 \times 10^{- 26}}} = 523 m/s$

Q 28 :

If the r.m.s. speed of chlorine molecule is 490 m/s at 27°C, the r.m.s. speed of argon molecules at the same temperature will be (Atomic mass of argon = 39.9 $u$ , molecular mass of chlorine = 70.9 $u$ ) [2023]

751.7 m/s
451.7 m/s
651.7 m/s
551.7 m/s

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(3)

$V_{rms} = \sqrt{\frac{3 R T}{M}}, \frac{v_{Ar}}{v_{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}}$

$\Rightarrow V_{Ar} = 1.33 \times 490 = 651.7 m/s$

Q 29 :

The r.m.s. speed of oxygen molecule in a vessel at a particular temperature is ${(1 + \frac{5}{x})}^{1 / 2} v,$ where $v$ is the average speed of the molecule. The value of $x$ will be: (Take $π = \frac{22}{7}$ ) [2023]

28
27
8
4

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4

(1)

$\sqrt{\frac{3 R T}{M}} = {(1 + \frac{5}{x})}^{\frac{1}{2}} \sqrt{\frac{8 R T}{π M}}$

$\Rightarrow \frac{3 \times 22}{7 \times 8} = 1 + \frac{5}{x} \Rightarrow x = 28$

Q 30 :

The mean free path of molecules of a certain gas at STP is 1500 $d$ , where $d$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately [2023]

1098 $d$
2049 $d$
750 $d$
1500 $d$

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(2)

$Mean free path, λ = \frac{R T}{\sqrt{2} π d^{2} N_{A} P}$

$λ \propto T$

$\frac{1500 d}{λ} = \frac{273}{373} \Rightarrow λ = 2049 d$

Topic Question Set

Q 21 :

At 300 K, the rms speed of oxygen molecules is $\sqrt{\frac{α + 5}{α}}$ times to that of its average speed in the gas. Then, the value of $α$ will be (use $π = \frac{22}{7}$ ) [2023]

Q 22 :

Heat energy of 735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but does not oscillate. The increase in the internal energy of the gas will be [2023]

Q 23 :

The average kinetic energy of a molecule of the gas is [2023]

Q 24 :

The temperature of an ideal gas is increased from 200 K to 800 K. If the r.m.s. speed of the gas at 200 K is $v_{0}$ , then the r.m.s. speed of the gas at 800 K will be [2023]

Q 27 :

The root mean square speed of molecules of nitrogen gas at $27 ° C$ is approximately:

(Given mass of a nitrogen molecule $= 4.6 \times 10^{- 26} kg$ and take Boltzmann constant $k_{B} = 1.4 \times 10^{- 23} {JK}^{- 1}$ ) [2023]

Q 28 :

If the r.m.s. speed of chlorine molecule is 490 m/s at 27°C, the r.m.s. speed of argon molecules at the same temperature will be (Atomic mass of argon = 39.9 $u$ , molecular mass of chlorine = 70.9 $u$ ) [2023]

Q 29 :

The r.m.s. speed of oxygen molecule in a vessel at a particular temperature is ${(1 + \frac{5}{x})}^{1 / 2} v,$ where $v$ is the average speed of the molecule. The value of $x$ will be: (Take $π = \frac{22}{7}$ ) [2023]

Q 30 :

The mean free path of molecules of a certain gas at STP is 1500 $d$ , where $d$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately [2023]

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Topic Question Set

Q 23 : The average kinetic energy of a molecule of the gas is [2023] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 23 :

The average kinetic energy of a molecule of the gas is [2023]