Work done on heating one mole of monoatomic gas adiabatically through $20^{\circ} C$ is $W$. Then,the work done on heating $6$ moles of rigid diatomic gas through the same change in temperature is: (in $W$)

The relation between volume $(V)$ and absolute temperature $(T)$ of a gas in an adiabatic process 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

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

$A$ 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 nitrogen gas being initially at a temperature of $T_0 = 300 \,K$ is adiabatically compressed to increase its pressure $10$ times. The final gas temperature after compression is (Assume,nitrogen gas molecules as rigid diatomic and $100^{1/7} = 1.9$) (in $\,K$)