Two vessels separately contain two ideal gases \(\mathrm{A}\) and \(\mathrm{B}\) at the same temperature, the pressure of \(\mathrm{A}\) being twice that of \(\mathrm{B}\). Under such conditions, the density of \(\mathrm{A}\) is found to be \(1.5\) times the density of \(\mathrm{B}\). The ratio of molecular weight of \(\mathrm{A}\) and \(\mathrm{B}\) is:

One mole of an ideal diatomic gas undergoes a transition from \(A\) to \(B\) along a path \(AB\) as shown in the figure. 
         
The change in internal energy of the gas during the transition is:

The mean free path of molecules of a gas (radius \(r\)) is inversely proportional to:
1. \(r^3\)
2. \(r^2\) 
3. \(r\) 
4. \(\sqrt{r}\)

In the given \({(V\text{-}T)}\) diagram, what is the relation between pressure \({P_1}\) and \({P_2}\)? 
              

The amount of heat energy required to raise the temperature of \(1\) g of Helium at NTP, from \({T_1}\) K to \({T_2}\) K is:
1. \(\frac{3}{2}N_ak_B(T_2-T_1)\)
2. \(\frac{3}{4}N_ak_B(T_2-T_1)\)
3. \(\frac{3}{4}N_ak_B\frac{T_2}{T_1}\)
4. \(\frac{3}{8}N_ak_B(T_2-T_1)\)

If \(C_P\) and \(C_V\) denote the specific heats (per unit mass) of an ideal gas of molecular weight \(M\) (where \(R\) is the molar gas constant), the correct relation is:
1. \(C_P-C_V=R\)
2. \(C_P-C_V=\frac{R}{M}\)
3. \(C_P-C_V=MR\)
4. \(C_P-C_V=\frac{R}{M^2}\)

At \(10^{\circ}\text{C}\) the value of the density of a fixed mass of an ideal gas divided by its pressure is \(x.\) At \(110^{\circ}\text{C}\) this ratio is: