The amount of work done in an adiabatic expansion from temperature $T$ to ${T_1}$ is

When a gas expands adiabatically, its volume is doubled while its absolute temperature is decreased by a factor of $2$. The value of the adiabatic constant $\gamma$ is

$A$ diatomic gas $\left(\gamma = \frac{7}{5}\right)$ is compressed adiabatically to volume $\frac{V_0}{32}$,where $V_0$ is its initial volume. The initial temperature of the gas is $T_i$ in Kelvin and the final temperature is $xT_i$ in Kelvin. The value of $x$ is:

The pressure and density of a diatomic gas $(\gamma = 7/5)$ change adiabatically from $(P, d)$ to $(P', d')$. If $\frac{d'}{d} = 32$,then $\frac{P'}{P}$ is equal to:

In the following $P-V$ diagram, two adiabatics cut two isothermals at temperatures $T_1$ and $T_2$. The value of $\frac{V_a}{V_d}$ will be

$A$ certain volume of a gas at $300 \ K$ expands adiabatically until its volume is doubled. The resultant fall in temperature of the gas is nearly (The ratio of the specific heats of the gas $\gamma = 1.5$) (in $K$)