One mole of an ideal gas expands adiabatically at constant pressure such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\gamma$ for the gas is $(\gamma = \frac{C_p}{C_v}, V = \text{Volume of the gas})$

The internal energy of the gas increases in:

If $5.6$ litres of a monoatomic gas at $STP$ is adiabatically compressed to $0.7$ litres,then the work done on the gas is nearly ($R$ = Universal gas constant). (in $R$)

$A$ litre of dry air at $STP$ expands adiabatically to a volume of $3$ litres. If $\gamma=1.40,$ the work done by air is $(3^{1.4}=4.6555)$. [Take air to be an ideal gas] (in $; J$)

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 $(R=8.31 \ J \ mole^{-1} \ K^{-1})$.

One gram mole of a diatomic gas $(\gamma = 1.4)$ is compressed adiabatically so that its temperature rises from $27^{\circ}C$ to $127^{\circ}C$. The work done will be: