$A$ rigid diatomic gas $(\gamma = 7/5)$ is compressed adiabatically to volume $(V_i/32)$, where $V_i$ is the initial volume. The initial temperature of the gas is $T_i \text{ K}$ and the final temperature is $x T_i \text{ K}$. The value of $x$ 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:

When air of the atmosphere rises up,it cools. Why?

Consider a thermodynamic process where internal energy $U = A P^2 V$ $(A = \text{constant})$. If the process is performed adiabatically, then:

$5$ moles of Hydrogen $\left(\gamma=\frac{7}{5}\right)$ initially at $S.T.P.$ are compressed adiabatically so that its temperature becomes $400^{\circ} C$. The increase in the internal energy of the gas in kilo-joules is $\left(R=8.30 \ J \ mol^{-1} \ K^{-1}\right)$.

The volume of an ideal gas $(\gamma=1.5)$ is changed adiabatically from $5 \ L$ to $4 \ L$. The ratio of initial pressure to final pressure is: