A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

A poly-atomic molecule $\left(C_V=3 R, C_P=4 R\right.$, where $R$ is gas constant) goes from phase space point $\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.$ $\mathrm{Pa}, \mathrm{V}_C=8 \times 10^{-6} \mathrm{~m}^3$ ). A to $B$ is an adiabatic path and $B$ to $C$ is an isothermal path.

The net heat absorbed per unit mole by the system is :

The work done in an adiabatic change in an ideal gas depends upon only :

The ratio of vapour densities of two gases at the same temperature is $ \frac{4}{25} $, then the ratio of r.m.s. velocities will be :