In a reversible isochoric change,

An ideal gas undergoes the process $1 \rightarrow 2$ as shown in the figure. The heat supplied and work done in the process are $\Delta Q$ and $\Delta W$ respectively. The ratio $\Delta Q : \Delta W$ is

$A$ diatomic gas $(\gamma=1.4)$ does $100 \,J$ of work in an isobaric expansion. The heat given to the gas is: (in $\,J$)

$A$ container having $1 \ mol$ of a gas at a temperature of $27 \ ^oC$ has a movable piston which maintains a constant pressure of $1 \ atm$ in the container. The gas is compressed until the temperature becomes $127 \ ^oC$. The work done is ........ $J$ $(R = 8.31 \ J/mol-K)$.

$10 \text{ mole}$ of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ at constant volume $V = 1 \text{ m}^3$ is $\alpha \text{ Joule}$ ($P_1 = 21.7 \text{ Pa}$, $P_2 = 30 \text{ Pa}$, $C_v = 21 \text{ J/K} \cdot \text{mol}$, $R = 8.3 \text{ J/mol} \cdot \text{K}$). The value of $\alpha$ is . . . . . . .

Two moles of helium gas are taken over the cycle $ABCDA$,as shown in the $P-T$ diagram. Assuming the gas to be ideal,the work done on the gas in taking it from $A$ to $B$ is ...... $R$.