$A$ given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be $0.75$ of its initial pressure. The ratio of the specific heats of the gas is approximately:

$A$ cyclic process $ABCA$ is shown in the $V-T$ diagram. The corresponding process on the $P-V$ diagram is:

In changing the state of a gas adiabatically from an equilibrium state $A$ to another equilibrium state $B$,an amount of work equal to $22.3 \; J$ is done on the system. If the gas is taken from state $A$ to $B$ via a process in which the net heat absorbed by the system is $9.35 \; cal$,how much is the net work done (in $J$) by the system in the latter case? (Take $1 \; cal = 4.19 \; J$)

The efficiency of the cycle shown in the figure (consisting of one isobar,one adiabat,and one isotherm) is $50 \%$. The ratio $x$ between the highest and lowest temperatures attained in this cycle obeys (the working substance is an ideal gas):

Two gases $A$ and $B$ have the same initial state $(P, V, n, T)$. Gas $A$ is compressed to $V/8$ by an isothermal process,and gas $B$ is compressed to $V/8$ by an adiabatic process. The ratio of the final pressure of gas $A$ to that of gas $B$ is (Both gases are monoatomic,$\gamma = 5/3$).

$A$ 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):
$(A)$ $q_{AC} = \Delta U_{AC}$ and $W_{AB} = 0$
$(B)$ $W_{BC} = P_2(V_1 - V_2)$ and $q_{BC} = \Delta H_{BC}$
$(C)$ $\Delta H_{CA} < \Delta U_{CA}$ and $q_{AC} = \Delta U_{AC}$
$(D)$ $q_{BC} = \Delta H_{BC}$ and $\Delta H_{CA} > \Delta U_{CA}$