$540$ calories of heat converts $1$ cubic centimeter of water at $100\,^oC$ into $1671$ cubic centimeter of steam at $100\,^oC$ at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly .......... $cal$.

The temperature-entropy $(T-S)$ diagram of a reversible engine cycle is given in the figure. Its efficiency is

$A$ monatomic gas of volume $V$ and pressure $P$ expands isothermally to a volume $27 V$ and then is compressed adiabatically to a volume $V$. The final pressure of the gas is: (in $P$)

Two moles of a gas are expanded to double their volume by two different processes. One is isobaric and the other is isothermal. If $W_1$ and $W_2$ are the works done, respectively, then:

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 $, J$)?

One mole of a monoatomic ideal gas goes through a thermodynamic cycle,as shown in the volume versus temperature $(V-T)$ diagram. The correct statement$(s)$ is/are :
[$R$ is the gas constant]
$(1)$ Work done in this thermodynamic cycle $(1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1)$ is $|W| = \frac{1}{2} RT_0$
$(2)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $2 \rightarrow 3$ is $\left|\frac{Q_{1 \rightarrow 2}}{Q_{2 \rightarrow 3}}\right| = \frac{5}{3}$
$(3)$ The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
$(4)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $3 \rightarrow 4$ is $\left|\frac{Q_{1 \rightarrow 2}}{Q_{3 \rightarrow 4}}\right| = \frac{1}{2}$