One mole of a diatomic ideal gas undergoes a process shown in $P-V$ diagram. The total heat given to the gas $(\ln 2 = 0.7)$ is (in $P_0 V_0$)

Two cylinders $A$ and $B$ are fitted with pistons and contain equal amounts of a diatomic gas at $300 \ K$. The piston of cylinder $A$ is free to move,while the piston of cylinder $B$ is fixed. If the same amount of heat is supplied to each cylinder,the temperature of gas in $A$ increases by $30 \ K$. What is the increase in the temperature of gas in $B$ (in $K$)?

The specific heat at constant pressure of a real gas obeying $PV^2=RT$ equation is:

$A$ thermodynamic system goes from states $(i) \, P_1, V$ to $2P_1, V$ and $(ii) \, P, V$ to $P, 2V$. The work done in the two cases is:

Five moles of an ideal gas has pressure $p_0$, volume $V_0$, and temperature $T_0$. The gas is expanded to volume $3V_0$ along a path such that the pressure $p$ changes as a function of volume $V$ as $p = p_0(V/V_0)$. The pressure is then reduced to $p_0$ while maintaining constant volume. Finally, the gas undergoes an isobaric compression until the volume and temperature return to $V_0$ and $T_0$, respectively. The total work done by the gas during the entire process is:

Three processes form a thermodynamic cycle as shown on the $P-V$ diagram for an ideal gas. Process $1 \rightarrow 2$ takes place at a constant temperature $(300 \ K)$. Process $2 \rightarrow 3$ takes place at a constant volume. During this process,$40 \ J$ of heat leaves the system. Process $3 \rightarrow 1$ is adiabatic and the temperature $T_3$ is $275 \ K$. The work done by the gas during the process $3 \rightarrow 1$ is ..... $J$.