Consider a gas with density $\mathrm{\rho }$ and $\overline{\mathrm{c}}$ as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity v, then the pressure exerted by the gas is

1.  $\frac{1}{3}\mathrm{\rho }{\overline{\mathrm{c}}}^{ 2}$                                   

2.  $\frac{1}{3}\mathrm{\rho }{\left(\mathrm{c}+\mathrm{v}\right)}^2$

3.  $\frac{1}{3}\mathrm{\rho }{\left(\overline{\mathrm{c}}-\mathrm{v}\right)}^2$                         

4.  $\frac{1}{3}\mathrm{\rho }{\left({\mathrm{c}}^{-2}-\mathrm{v}\right)}^2$

If the mean free path of atoms is doubled, then the pressure of the gas will become:

1. \(\frac{P}{4}\)                   
2. \(\frac{P}{2}\)
3. \(\frac{P}{8}\)
4. \(P\)

A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from 300 K to 600 K, the average K.E. of helium atoms is

1. Halved                     

2. Unchanged

3. Doubled                   

4. Increased by factor $\sqrt{2}$

The kinetic energy of translation of 20 gm of oxygen at 47°C is (molecular wt. of oxygen is 32 gm/mol and R = 8.3 J/mol/K)

1.  2490 joules                   

2.  2490 ergs

3.  830 joules                     

4.  124.5 joules

The kinetic energy of one gram molecules of a gas at standard temperature and pressure: $\left(\mathrm{R}=8.31 \mathrm{J}/\mathrm{mol}-\mathrm{K}\right)$

1.  $0.56\times {10}^4 \mathrm{J}$                           

2.  $1.3\times {10}^2 \mathrm{J}$

3.  $2.7\times {10}^2 \mathrm{J}$                             

4.  $3.4\times {10}^3 \mathrm{J}$

A jar contains a gas and few drops of water at T K. The pressure in the jar is 830 mm of mercury. The temperature of jar is reduced by 1%. The saturated vapour pressure of water at the two temperatures are 30 mm and 25 mm of mercury. Then the new pressure in the jar will be

1.  917 mm of Hg                       

2.  717 mm of Hg

3.  817 mm of Hg                       

4.  None of these

Assertion : At a particular temperature, the value of mean free path increases with decrease in pressure.

Reason : All the gas molecules at a particular temperature possess same speed.

The relation between two specific heats (in cal/mol) of a gas is:
1.  ${\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{J}}$                               

2.  ${\mathrm{C}}_{\mathrm{V}}-{\mathrm{C}}_{\mathrm{P}}=\frac{\mathrm{R}}{\mathrm{J}}$

3.  ${\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}=\mathrm{J}$                                 

4.  ${\mathrm{C}}_{\mathrm{V}}-{\mathrm{C}}_{\mathrm{P}}=\mathrm{J}$

Gas at a pressure ${\mathrm{P}}_0$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to

1.  $4{\mathrm{P}}_0$                                 

2.  $2{\mathrm{P}}_0$

3.  ${\mathrm{P}}_0$                                   

4.  $\frac{{\mathrm{P}}_0}{2}$

The specific heats at constant pressure is greater than that of the same gas at constant volume because

1. At constant pressure work is done in expanding the gas

2. At constant volume work is done in expanding the gas

3. The molecular attraction increases more at constant pressure

4. The molecular vibration increases more at constant pressure