An ideal monoatomic gas is carried along the cycle $ABCDA$ as shown in the figure. The total heat absorbed during this process is (in $p_0 V_0$)

An ideal gas at initial temperature $T_0$ and initial volume $V_0$ is expanded adiabatically to a volume $2 V_0$. The gas is then expanded isothermally to a volume $5 V_0$ and thereafter compressed adiabatically so that the temperature of the gas becomes again $T_0$. If the final volume of the gas is $\alpha V_0$, then the value of constant $\alpha$ is

Suppose an ideal gas ($n$ moles) undergoes an expansion process $P = f(V)$ which passes through the point $(V_0, P_0)$. If the slope of the curve $P = f(V)$ is greater than the slope of the adiabatic curve passing through $(V_0, P_0)$,show that the gas absorbs heat at $(V_0, P_0)$.

In a certain thermodynamical process,the pressure of a gas depends on its volume as $P = kV^{3}$. The work done when the temperature changes from $100^{\circ}C$ to $300^{\circ}C$ will be .......... $nR$,where $n$ denotes the number of moles of a gas.

One mole of an ideal diatomic gas expands from volume $V$ to $2V$ isothermally at a temperature $27^{\circ} C$ and does $W$ joule of work. If the gas undergoes the same magnitude of expansion adiabatically from $27^{\circ} C$ doing the same amount of work $W$, then its final temperature will be (close to) . . . . . . ${ }^{\circ} C$.

Two identical vessels $A$ and $B$ contain equal amounts of an ideal monoatomic gas. The piston of $A$ is fixed,while the piston of $B$ is free to move. The same amount of heat $\Delta Q$ is absorbed by both $A$ and $B$. If the internal energy of $B$ increases by $100 \, J$,what is the change in the internal energy of $A$ (in $, J$)?