Question

For an ideal gas, an illustration of three different paths $A$, $(B+C)$ and $(D+E)$ from an initial state $P_1$, $V_1$, $T_1$ to a final state $P_2$, $V_2$, $T_1$ is shown in the given figure:

Path A represents a reversible Isothermal expansion from $P_1$,$V_{1}to{P}_2,V_2$. Path $(B+C)$ reperesents a reversible adiabatic expansion $\left(B\right)$ from $P_1{V}_1$, $T_{1}to{P}_3$, $V_2$, $T_2$ followed by reversible heating the gas at constant volume (C) from $P_3$, $V_2$, $T_{2}to{P}_2$, $V_2$, $T_1$. Path ?$(D+E)$ represents a reversible expansion at constant pressure $P_1\left(D\right)$ from $P_1$, $V_1$, $T_1$ to $P_1$, $V_2$, $T_3$ followed by a reversible cooling at constant volume $V_2\left(E\right)$ from $P_1$, $V_2$, $T_{3}to{P}_2$, $V_2$, $T_1$

What is $q_{rev}$, for path $(B+C)?$

• A. $nRln\frac{V_2}{V_1}$
• B. $-nRln\frac{V_2}{V_1}$
• C. $zero$
• D. $nR{T}_{1}ln\frac{V_2}{V_1}$

Answer 1

The correct option is B $-nRln\frac{V_2}{V_1}$

Path (B+C) occurs in adiabatic conditions, hence $q=0$

We know that $\Delta S=\frac{q_{rev}}{T}{q}_{rev}=T\Delta S$

$q_{rev}$ = $w$

$q_{rev}$ =0 + (- $nRln\frac{V_2}{V_1}$)

$q_{rev}=-nRln\frac{V_2}{V_1}$