$A$ gas at $N.T.P.$ is suddenly compressed to one-fourth of its original volume. If $\gamma = 1.5$,then the final pressure is

$A$ van der Waals gas obeys the equation of state $\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$. Its internal energy is given by $U=C T-\frac{n^2 a}{V}$. The equation of a quasistatic adiabat for this gas is given by

$A$ sample of $1$ mole of gas at temperature $T$ is adiabatically expanded to double its volume. If the adiabatic constant for the gas is $\gamma = \frac{3}{2}$,then the work done by the gas in the process is:

In a thermodynamic system,$\Delta U$ represents the increase in its internal energy and $dW$ is the work done by the system. Then,which of the following statements is correct?

One mole of an ideal gas with $\gamma = 1.4$ is adiabatically compressed so that its temperature rises from $27^{\circ}C$ to $35^{\circ}C$. The change in the internal energy of the gas is ....... $J$ $(R = 8.3 \, J/mol \cdot K)$.

The work of $146 \ kJ$ is performed in order to compress one kilo mole of gas adiabatically,and in this process,the temperature of the gas increases by $7 ^\circ C$. The gas is $(R = 8.3 \ J \ mol^{-1} K^{-1})$.