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