The change in the internal energy of an ideal gas does not depend on?

1.	Number of moles
2.	Change in temperature
3.	Specific heat at constant pressure \(C_p\) of the gas
4.	Specific heat at constant volume \(C_v\) of the gas

When the gas in an open container is heated, the mean free path:

The translational kinetic energy of oxygen molecules at room temperature is 60 J. Their rotational kinetic energy will be?

1. 40 J

2. 60 J

3. 50 J

4. 20 J

Two isotherms are drawn at temperatures ${\mathrm{T}}_1$ $\mathrm{and}$ ${\mathrm{T}}_2$ as shown. The ratio of mean speed  at ${\mathrm{T}}_1$ $\mathrm{and}$ ${\mathrm{T}}_2$ is:
  

The translational kinetic energy of n moles of a diatomic gas at absolute temperature T is given by:
1. $\frac{5}{2}nRT$
2. $\frac{3}{2}nRT$
3. $\mathit{5}nRT$
4. $\frac{7}{2}nRT$

If the pressure of a gas is doubled, then the average kinetic energy per unit volume of the gas will be:

The figure shows a process for a gas in which pressure (P) and volume (V) of the gas change. If $C_1$ and $C_2$ are the molar heat capacities of the gas during the processes AB and BC respectively, then:

1. $C_1=C_2$

2. $C_1>C_2$

3. $C_1<C_2$

4. $C_1\le C_2$

In the PV graph shown below for an ideal diatomic gas, the change in the internal energy is:
     

1. $\frac{3}{2}\mathrm{P}({\mathrm{V}}_2-{\mathrm{V}}_1)$

2. $\frac{5}{2}\mathrm{P}({\mathrm{V}}_2-{\mathrm{V}}_1)$

3. $\frac{3}{2}\mathrm{P}({\mathrm{V}}_1-{\mathrm{V}}_2)$

4. $\frac{7}{2}\mathrm{P}({\mathrm{V}}_1-{\mathrm{V}}_2)$

How does the pressure of an ideal gas change during the process shown in the diagram?
       

The pressure in a diatomic gas increases from ${\mathrm{P}}_0$ to $3{\mathrm{P}}_0$, when its volume is increased from ${\mathrm{V}}_0$ $\mathrm{to}$ $2{\mathrm{V}}_0$. The increase in internal energy  will be:
     
1. $6{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$

2. $8.5{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$

3. $12.5{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$

4. $14.5{\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_0$