An ideal gas at pressure $P_0$ undergoes an isothermal expansion until its volume is $8.0$ times its initial volume. The gas is slowly and adiabatically compressed back to its original volume. If the adiabatic constant of the gas is $\gamma = 4/3$, then the ratio of the average kinetic energy per molecule in this final state to that in the initial state is

The ratio of the slopes of isothermal and adiabatic curves is

$A$ reversible cyclic process for an ideal gas is shown below. Here,$P, V$,and $T$ are pressure,volume,and temperature,respectively. The thermodynamic parameters $q, w, H$,and $U$ are heat,work,enthalpy,and internal energy,respectively.
The correct option$(s)$ is (are):
$(A)$ $q_{AC} = \Delta U_{BC}$ and $W_{AB} = P_2(V_2 - V_1)$
$(B)$ $W_{BC} = P_2(V_2 - V_1)$ and $q_{BC} = H_{AC}$
$(C)$ $\Delta H_{CA} < \Delta U_{CA}$ and $q_{AC} = \Delta U_{BC}$
$(D)$ $q_{BC} = \Delta H_{AC}$ and $\Delta H_{CA} > \Delta U_{CA}$

The figure shows the $P-V$ diagram of a thermodynamic cycle. Which corresponding curve is correct?

If $R$ is the universal gas constant,the amount of heat needed to raise the temperature of $2$ moles of an ideal monoatomic gas from $273 \ K$ to $373 \ K$ when no work is done is equal to ...... $R$.

$A$ heating element of resistance $r$ is fitted inside an adiabatic cylinder which carries a frictionless piston of mass $m$ and cross-sectional area $A$. The cylinder contains one mole of a diatomic gas. The temperature of the gas varies with time $t$ as $T = \alpha t + \frac{1}{2} \beta t^2$ (where $\alpha$ and $\beta$ are constants), while the pressure remains constant. The atmospheric pressure above the piston is $P_0$. Then: