$A$ rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for the process is $TV^x =$ constant,then $x$ is

One mole of helium is adiabatically expanded from its initial state $({P_i}, {V_i}, {T_i})$ to its final state $({P_f}, {V_f}, {T_f})$. The decrease in the internal energy associated with this expansion is equal to

The temperature of a hypothetical gas increases to $\sqrt{2}$ times when compressed adiabatically to half the volume. Its equation can be written as

When a gas is compressed adiabatically by a piston in a cylinder,the work done on the gas is found to be $1.5 \times 10^{4} \, J$. During this process:

At $N.T.P.$,one mole of a diatomic gas is compressed adiabatically to half of its volume. Given $\gamma = 1.41$,the work done on the gas will be approximately ....... $J$.

$A$ gas at normal temperature is suddenly compressed to one-fourth of its original volume. If $\gamma = 1.5$, then the increase in the temperature of the gas in Kelvin is ($\gamma$ is the ratio of specific heats).