Two thermally insulated vessels \(1\) and \(2\) are filled with air at temperatures \(\mathrm{T_1},\) \(\mathrm{T_2},\) volume \(\mathrm{V_1},\) \(\mathrm{V_2}\) and pressure \(\mathrm{P_1},\) \(\mathrm{P_2}\) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be:
1. \(T_1+T_2\)

2. \(\frac{T_1+T_2}{2}\)

3. \(\frac{T_1T_2(P_1V_1+P_2V_2)}{P_1V_1T_2+P_2V_2T_1}\)

4. \(\frac{T_1T_2(P_1V_1+P_2V_2)}{P_1V_1T_1+P_2V_2T_2}\)

If \(V_\text{H}\),$$\(V_\text{N}\) and \(V_\text{O}\) denote the root-mean square velocities of molecules of hydrogen, nitrogen and oxygen respectively at a given temperature, then:
1. \(V_\text{N}>V_\text{O}>V_\text{H}\)
2. \(V_\text{H}>V_\text{N}>V_\text{O}\)
3. \(V_\text{O}>V_\text{N}>V_\text{H}\)
4. \(V_\text{O}>V_\text{H}>V_\text{N}\)

If the ratio of vapour density for hydrogen and oxygen is \(1 \over 16\), then under constant pressure, the ratio of their rms velocities will be:

What is the velocity of a wave in a monoatomic gas having pressure 1 kilopascal and density $2.6 \mathrm{kg}/{\mathrm{m}}^3$ ?

1.  $3.6 \mathrm{m}/\mathrm{s}$                                 

2.  $8.9\times {10}^3 \mathrm{m}/\mathrm{s}$

3.  Zero                                       

4.  None of these

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

1. P/4                       

2. P/2

3. P/8                       

4. P

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

The rms speed of oxygen atoms is v. If the temperature is halved and the oxygen atoms combine to form oxygen molecules, then the rms speed will be:

1. $\frac{v}{\sqrt{2}}$

2. $v\sqrt{2}$

3. 2v

4. $\frac{v}{2}$

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

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

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