Thermodynamics and Thermochemistry - Result Question 68

68. Two moles of a perfect gas undergo the following processes :

(a) a reversible isobaric expansion from $(1.0 $ $atm, 20.0 L)$ to $(1.0 $ $atm, 40.0 L)$

(b) a reversible isochoric change of state from $(1.0 $ $atm$, $40.0 L)$ to $(0.5 $ $atm, 40.0 L)$

(c) a reversible isothermal compression from $(0.5 $ $atm$, $40.0 L)$ to $(1.0 $ $atm, 20.0 L)$

(i) Sketch with labels each of the processes on the same $p$ - $V$ diagram.

(ii) Calculate the total work $(W)$ and the total heat change $(Q)$ involved in the above processes.

(iii) What will be the values of $\Delta U, \Delta H$ and $\Delta S$ for the overall process?

(2002, 5M)

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Solution:

(i)

(ii)

$ \begin{aligned} -W _1 & =p \Delta V=20 L \text { atm } \\ W _2 & =0 \quad \because \quad \Delta V=0 \\ W _3 & =n R T \ln \frac{40}{20}=20 \ln 2 \end{aligned} $

Total work done $=W _1+W _2+W _3$

$ \begin{aligned} & =-20 \hspace{1mm} L \hspace{1mm} atm+0+20 \ln 2 \\ & =-6.14 \hspace{1mm} atm \end{aligned} $

From first law : $q=\Delta E+(-W)=-W$

($\therefore \Delta E = 0$ for cyclic process)

$ \Rightarrow \quad q=6.14 $ $L$ $ atm=622.53 J $

(iii) All the states function, $\Delta U, \Delta H$ and $\Delta S$ are zero for cyclic process.