Question

Two moles of a perfect gas undergo the following processes
I. A reversible isobaric expansion from (1.0 atm. 20 L) to (1.0 atm, 40 L)
II. A reversible isochoric exansion from (1.0 atm, 40 L) to (0.5 atm, 40 L)
III. A reversible isothermal compression from (0.5 atm, 40 L) to (1.0 atm. 20 L)
Calculate the total work done (W) involved in the above processes.

• A. -620.77 J
• B. 620.77 J
• C. -120.77 J
• D. 120.77 J

Answer 1

The correct option is A -620.77 J

The given 3 processes comprises a cyclic system ABC which is shown in the graph below.

Total workdone in cyclic process, $W=W_{A\to B}+W_{B\to C}+W_{C\to A}$

Workdone in $A\to B$ process :
It is a isobaric expansion (at constant pressure, P = 1 atm)
Initial volume, $V_A$= 20 L
Final volume, $V_B$ = 40 L
Work done, $W_{A\to B}=-P(V_B-V_A)$
$=-1(40-20)=-20L.atm$

Workdone in $B\to C$ process :
It is a reversible isochoric process. The volume is constant at 40 L so there is zero workdone during this process.
$W_{B\to C}=0$

Workdone in $C\to A$ process :
It is an isothermal compression (at constant temperature,T) process.
Initial volume, $V_C$= 20 L
Final volume, $V_A$ = 40 L
Wordone for isothermal compression,
$W_{C\to A}=2.303nRTlog\frac{V_C}{V_A}=2.303P_A{V}_A\log \frac{40}{20}=13.87L.atm$

Total workdone in cyclic process, $W=W_{A\to B}+W_{B\to C}+W_{C\to A}$
$W=-20+0+13.87=-6.13L.atm=-\frac{6.13\times 8.314}{0.0821}J\left(injoules\right)=-620.77J$