One mole of an ideal gas expands adiabatically from $200 \,K$ to $250 \,K$. If the specific heat of the gas at constant volume is $0.8 \,kJ \,kg^{-1} \,K^{-1}$, then the work done by the gas is

In a thermodynamic process, helium gas obeys the law $TP^{-2/5} = \text{constant}$. The heat given to the gas when the temperature of $2$ moles of the gas is raised from $T$ to $4T$ ($R$ is the universal gas constant) is:

Two moles of an ideal monoatomic gas at $27^{\circ}C$ occupies a volume of $V$. If the gas is expanded adiabatically to the volume $2V$,then the work done by the gas will be ....... $J$ $[\gamma = 5/3, R = 8.31 \text{ J/mol K}]$

The phenomenon of sound propagation in air is

$A$ spherical bubble inside water has radius $R$. Take the pressure inside the bubble and the water pressure to be $p_0$. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes $(R-a)$. For $a \ll R$,the magnitude of the work done in the process is given by $(4 \pi p_0 R a^2) X$,where $X$ is a constant and $\gamma = C_p / C_V = 41 / 30$. The value of $X$ is:

During an adiabatic process,the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $\frac{C_P}{C_V}$ for the gas is