$A$ certain volume of a gas at $300 \ K$ expands adiabatically until its volume is doubled. The resultant fall in temperature of the gas is nearly (The ratio of the specific heats of the gas $\gamma = 1.5$) (in $K$)

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 ....... $J$ $(R = 8.3 \, J/mol \cdot K)$.

$A$ 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

$A$ monatomic gas at a pressure of $100 \text{ kPa}$ expands adiabatically such that its final volume becomes $8$ times its initial volume. If the work done during the process is $180 \text{ J}$, then the initial volume of the gas is (in $\text{ cm}^3$)

The $P-V$ diagram for two adiabatic processes is given. Curves $1$ and $2$ correspond to:

The work done in which of the following processes is equal to the internal energy of the system?