During an adiabatic process, the pressure is proportional to the cube of the temperature. Then ${C_p}/{C_v}$ is . . . . . . .

$5.6\, L$ of helium gas at $STP$ is adiabatically compressed to $0.7\, L$. Taking the initial temperature to be $T_1$,the magnitude of work done in the process is

An ideal gas expands adiabatically $(\gamma = 1.5)$. To reduce the r.m.s. velocity of the molecules $3$ times, the gas has to be expanded by a factor of: (in $\times$)

During an adiabatic process,the pressure of the gas is found to be proportional to the cube of its absolute temperature. The ratio $C_P/C_V = \gamma$ for the gas is:

An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes $32$ times its initial value. If the final pressure of the gas is $128$ atmospheres,the value of $\gamma$ for the gas is:

The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ change adiabatically from $(P, \rho)$ to $(P^{\prime}, \rho^{\prime})$. If $\frac{\rho^{\prime}}{\rho}=32$,then $\frac{P^{\prime}}{P}$ is: