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The root mean square speed of the molecules of an enclosed gas is V. What will be the root mean square speed if the pressure is doubled, the temperature remaining the same?

$\frac{v}{2}$

4 v

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Explanation

Root mean square speed is independent of pressure if the temperature remains the same.

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If the degree of freedom of gas are f, then the ratio of two specific heats $C_{P} / C_{V}$ is given by:

[MP PET 1995; BHU 1997; MP PMT 2001, 04]

$\frac{2}{f} + 1$

$1 - \frac{2}{f}$

$1 + \frac{1}{f}$

$1 - \frac{1}{f}$

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Explanation

(a)

$\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f}$

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The equation $(p + \frac{a}{v^{2}}) (v - b) = R T$ is known as:

Perfect gas equation

Joule Thomson's equation

Vander Waal's equation

Maxwell's equation

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Explanation

(c)

$(P + \frac{a}{V^{2}}) (V - b)$ is Vander Waal's gas equation of state.

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The temperature of an ideal gas is increased from $27 °$ to $927 ° C$ . The r.m.s. speed of its molecules becomes-

twice

half

four times

one fourth

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Explanation

(a)

$\frac{V_{r m s} (927 ° C)}{V_{r m s} (27 ° C)} = \sqrt{\frac{273 + 927}{273 + 27}}$

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An ideal gas is filled in a vessel, then

If it is placed inside a moving train, its temperature increases

Its centre of mass moves randomly

Its temperature remains constant in a moving car

None of these

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Explanation

(C)

The temperature depends on internal kinetic energy due to the random motion of gas molecules. It will not depend on the speed of the train.

Though bulk kinetic energy of gas will increase.

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If the molecular weight of two gases are $M_{1} a n d M_{2}$ , then at a temperature the ratio of root mean square velocity $v_{1} a n d v_{2}$ will be: [MP PMT 1996; AMU (Engg.) 2000; CPMT 2000; DPMT 2001]

$\sqrt{\frac{M_{1}}{M_{2}}}$

$\sqrt{\frac{M_{2}}{M_{1}}}$

$\sqrt{\frac{M_{1} + M_{2}}{M_{1} - M_{2}}}$

$\sqrt{\frac{M_{1} - M_{2}}{M_{1} + M_{2}}}$

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Explanation

(b)

$v_{r m s} = \sqrt{3 R T / M} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

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An ideal gas is found to obey an additional law $V P^{2}$ = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2 V, the temperature becomes:

T $\sqrt{2}$

2T $\sqrt{2}$

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Explanation

(a)

$V P^{2} = c o n s \tan t, P V = R T ∴ V \frac{R^{2} T^{2}}{V^{2}} = c o n s \tan t ∴ \frac{T^{2}}{V} = c o n s \tan t \frac{T^{2}}{V} = \frac{T'^{2}}{2 V} = T'^{2} = 2 T^{2} = T' = \sqrt{2} T$

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The root-mean-square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root mean square speed of oxygen molecules at 900 K will be: [MH CET 2002; Odisha JEE 2010]

1930 $\sqrt{3}$ m/s

836 m/s

643 m/s

$\frac{1930}{\sqrt{3}}$ m/s

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Explanation

(b)

$V_{r m s} = \sqrt{\frac{3 R T}{M}} \Rightarrow \frac{{(V_{r m s})}_{O_{2}}}{{(V_{r m s})}_{H_{2}}} = \sqrt{\frac{T_{O_{2}}}{T_{H_{2}}} \times \frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt{\frac{900}{300} \times \frac{2}{32}} {(V_{r m s})}_{O_{2}} = 836 m / s$

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What is the mass of 2 liters of nitrogen at 22.4 atmospheric pressure and 273 K?

28 g

14 $\times 22.4$ g

56 g

None of these

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Explanation

(c)

22.4 litres of nitrogen at 1 atmosphere or 1 litre at 22.4 atmospheric pressure has a mass of 1 gm mole or 28 g. Therefore, 2 litres at 22.4 atmospheres have a mass of 56 gm.

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Mean kinetic energy per degree of freedom of gas molecules is: [ MP PET 1995; RPET 1999, 2001, 03; Haryana CEET 2000; DCE 2009]

$\frac{3}{2} k T$

$\frac{1}{2} k T$

$\frac{3}{2} R T$

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Explanation

(c)

According to the law of equipartition of energy, kinetic energy per degree of freedom of a gas molecule is $\frac{1}{2} k T$ .

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