$A$ monoatomic gas at pressure $P$ having volume $V$ expands isothermally to a volume $2V$ and then adiabatically to a volume $16V$. The final pressure of the gas is $\left(\gamma = \frac{5}{3}\right)$.

Heat is supplied to a diatomic gas at constant pressure. The ratio of $\Delta Q: \Delta U: \Delta W$ is
[Given $\rightarrow \Delta Q=$ heat supplied,$\Delta U=$ change in internal energy,$\Delta W$ $=$ work done]

Two containers $A$ and $B$ contain equal volumes of an identical gas at the same pressure and temperature. The gas in container $A$ is compressed to half its original volume isothermally,while the gas in container $B$ is compressed to half its original volume adiabatically. The ratio of the final pressure of gas in container $B$ to that of gas in container $A$ 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 $\Delta H_{CA} > \Delta U_{CA}$

An office room contains about $2000$ moles of air. The change in the internal energy of this much air when it is cooled from $34^{\circ} C$ to $24^{\circ} C$ at a constant pressure of $1.0 \text{ atm}$ is (Use $\gamma_{\text{air}} = 1.4$ and universal gas constant $R = 8.314 \text{ J/mol-K}$)

In the $P-V$ diagram shown,the gas does $5 \, J$ of work in the isothermal process $ab$ and $4 \, J$ of work in the adiabatic process $bc$. What will be the change in internal energy of the gas in the straight path $c$ to $a$ (in $, J$)?