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:

$5.6\, L$ 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

The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$,then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$.

$A$ motor tube is filled with air at $27^{\circ}C$ and a pressure of $8 \text{ atm}$. If the tube suddenly bursts,what will be the final temperature of the air? $(\gamma = 1.5)$

One mole of an ideal gas at an initial temperature of $T$ $K$ does $6R$ of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5/3$,the final temperature of the gas will be $(R=8.31 \ J \ mole^{-1} \ K^{-1})$.

For an adiabatic compression of $1 \text{ kmol}$ of gas,$146 \text{ kJ}$ of work is done. During this process,the temperature of the gas increases by $7 \text{ °C}$. The gas is ........