$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given,$\gamma = 3/2$):

$Assertion :$ Adiabatic expansion is always accompanied by fall in temperature.
$Reason :$ In adiabatic process, volume is inversely proportional to temperature.

An engine takes in $5$ moles of air at $20\,^{\circ}C$ and $1\,atm$,and compresses it adiabatically to $1/10^{\text{th}}$ of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules,the change in its internal energy during this process is $X\,kJ$. The value of $X$ to the nearest integer is

The mean kinetic energy of monoatomic gas molecules under standard conditions is $\langle E_1 \rangle$. If the gas is compressed adiabatically $8$ times to its initial volume, the mean kinetic energy of gas molecules changes to $\langle E_2 \rangle$. The ratio $\frac{\langle E_2 \rangle}{\langle E_1 \rangle}$ is

In an adiabatic process,the state of a gas is changed from $P_1, V_1, T_1$ to $P_2, V_2, T_2$. Out of the following relations,the correct one is

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas $(\gamma = 5/3)$ decreases from $60 \text{ K}$ to $50 \text{ K}$. The work done by the gas in the process is : (Take the universal gas constant as $R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}$) (in $\text{ J}$)