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For a gas, $\frac{R}{C_{V}} = 0.67$ . This gas is made up of molecules which are: [JIPMER 2001, 02]

Diatomic

Mixture of diatomic and polyatomic molecules

Monoatomic

Polyatomic

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Explanation

(c)

$C_{V} = \frac{R}{0.67} = 1.5 R = \frac{3}{2} R$

This is the value of $C_{V}$ for monoatomic gases.

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At constant temperature, on increasing the pressure of a gas by 5% its volume will decrease by: [MP PET 2002; Kerala 2006]

5.26 %

4.26 %

4.76 %

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Explanation

$P \propto \frac{1}{V} \Rightarrow \frac{V_{2}}{V 1} = \frac{P_{1}}{P_{2}} = \frac{100}{105} \Rightarrow V_{2} = \frac{100}{105} V_{1} = 0.953 V_{1} % c h a n g e i n v o l u m e = \frac{V_{1} - V_{2}}{V 1} \times 100 = \frac{V_{1} - 0.953 V_{1}}{V_{1}} \times 100 = 4.76 %$

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At what temperature, is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at 47 $°$ C:[MP PET 1997, 2000; RPET 1999; AIEEE 2002; J & K CET 2004; Kerala PET 2010]

20 K

80 K

- 73 K

3 K

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Explanation

$V_{r m s} = \sqrt{\frac{3 R T}{M}} \Rightarrow T \propto M [∵ V_{r m s}, R \to c o n s \tan t] \Rightarrow \frac{T_{H_{2}}}{T_{O_{2}}} = \frac{M_{H_{2}}}{M_{O_{2}}} \Rightarrow \frac{T_{H_{2}}}{(273 + 47)} = \frac{2}{32} \Rightarrow T_{H_{2}} = 20 K$

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The average kinetic energy of a gas molecule can be determined by knowing: [RPET 2000, MP PET 2010]

The number of molecules in the gas

The pressure of the gas only

The temperature of the gas only

None of the above is enough by itself

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Explanation

(c)

$K i n e t i c e n e r g y \propto T e m p e r a t u r e$

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One litre of gas A and two litres of gas B, both having the same temperature 100 $°$ C and the same pressure 2.5 bar will have the ratio of kinetic energies of their molecules as:

1:1

1:2

1:4

4:1

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Explanation

$∴ K E = \frac{3}{2} k T$ , Since both the gases are at the same temperature, their mean KE are equal. The ratio is 1:1

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On 0 $°$ C, the pressure measured by the barometer is 760 mm. What will be pressure at 100 $°$ C? [AFMC 2002]

760 mm

730 mm

780 mm

None of these

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Explanation

$P \propto T \Rightarrow \frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}} = \frac{(273 + 100)}{(273 + 0)} = \frac{373}{273} \Rightarrow P_{2} = \frac{760 \times 373}{273} = 1038 m m$

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By what percentage, should the pressure of a given mass of gas be increased, so as to decrease its volume by 10% at a constant temperature?

7.2 %

12.5%

11.1%

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Explanation

Let $V_{1}$ be initial volume and $P_{1}$ be initial pressure.

Final volume = $V_{2} = V_{1}$ $- \frac{10}{100} V_{1} = \frac{90}{100} V_{1}$

If $P_{2}$ is final pressure, then

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

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} = \frac{100}{90} \frac{P_{2}}{P_{1}} - 1 = \frac{10}{9} - 1 = \frac{1}{9} \frac{P_{2} - P_{1}}{P_{1}} = \frac{1}{9} \times 100 = 11.1 %$

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The ratio $\frac{C_{p}}{C_{V}} = γ$ for a gas. Its molecular weight is M. Its specification heat capacity at constant pressure is

$\frac{R}{γ - 1}$

$\frac{γ R}{γ - 1}$

$\frac{γ R}{M (γ - 1)}$

$\frac{γ R M}{(γ - 1)}$

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Explanation

$C_{P} - C_{V} = R D i v i d i n g w i t h C_{P}, w e g e t 1 - \frac{1}{γ} = \frac{R}{C_{P}} o r \frac{γ - 1}{γ} = \frac{R}{C_{P}} o r C_{P} = \frac{γ R}{γ - 1}$

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The root-mean-square velocity of the molecules in a sample of helium is $\frac{5}{7} t h$ of that of the molecules in a sample of hydrogen. If the temperature of the hydrogen gas is 0 $°$ C, that of the helium sample is about:

0 $°$ C

273 $°$ C

100 $°$ C

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Explanation

(a)

$v_{r m s} o f H e a t t ° C = \frac{5}{7} (v_{r m s} o f H_{2} a t 0 ° C) ∴ \sqrt{\frac{3 R (273 + t)}{M_{H e}}} = \frac{5}{7} \sqrt{\frac{3 R (273 + 0)}{M_{m}}} \frac{3 R (273 + t)}{4 \times 10^{- 3}} = \frac{25}{49} \frac{3 R (273)}{2 \times 10^{- 3}} t = 0 ° C$

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Four molecules have speeds 2 km/sec, 3 km/sec, 4 km/sec and 5 km/sec. The root mean square speed of these molecules (km/sec) is:

$\sqrt{\frac{54}{4}}$

$\sqrt{\frac{54}{2}}$

3.5

3 $\sqrt{3}$

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Explanation

(a)

$v_{r m s} = \sqrt{\frac{2^{2} + 3^{2} + 4^{2} + 5^{2}}{4}} = \sqrt{\frac{54}{4}} k m s^{- 1}$

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