Two moles of helium gas are taken over the cycle $ABCDA$, as shown in the $P-T$ diagram. The net work done on the gas in the cycle $ABCDA$ is ...... $R$.

Match the thermodynamic processes taking place in a system with the correct conditions. In the table: $\Delta Q$ is the heat supplied,$\Delta W$ is the work done and $\Delta U$ is change in internal energy of the system.

Process	Condition
$(I)$ Adiabatic	$(A) \Delta W = 0$
$(II)$ Isothermal	$(B) \Delta Q = 0$
$(III)$ Isochoric	$(C) \Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$
$(IV)$ Isobaric	$(D) \Delta U = 0$

When $80 \ J$ of heat is supplied to a gas at constant pressure,if the work done by the gas is $20 \ J$,then the ratio of the specific heat capacities of the gas is

Check whether the following statements are true or false:
$1.$ The change in internal energy $\Delta U = 0$ in a cyclic process.
$2.$ In an adiabatic process,temperature remains constant.
$3.$ The internal energy of a system during an isothermal process decreases.

An engine operates by taking $n$ moles of an ideal gas through the cycle $ABCDA$ shown in the figure. The thermal efficiency of the engine is: (Take $C_v = 1.5 R$,where $R$ is the gas constant)

Two moles of monoatomic gas is expanded from $(P_0, V_0)$ to $(P_0, 2V_0)$ under isobaric condition. Let $\Delta Q_1$,$\Delta W_1$,and $\Delta U_1$ be the heat given to the gas,the work done by the gas,and the change in internal energy,respectively. Now,the monoatomic gas is replaced by a diatomic gas,with other conditions remaining the same. The corresponding values in this case are $\Delta Q_2$,$\Delta W_2$,and $\Delta U_2$. Then: