Question

Diagram shows the variation in the internal energy U with the volume V of 2.0 moles of an ideal gas in cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K, respectively. Calculate the heat absorbed by the gas during the process.

• A. 400 R ln 2
• B. 500 R ln 2
• C. 700 R ln 2
• D. 800 R ln 2

Answer 1

Answer: (A)

In the process a to b and c to d
as $\Delta \text{U}=0$ , therefore $\Delta \text{T}=0$ or T = constant
$\text{W}={\int }_{{\text{V}}_{\text{i}}}^{{\text{V}}_f}\text{PdV}$
We have,
$\text{PV}=\text{nRT}\Rightarrow \text{P}=\frac{\text{nRT}}{\text{V}}$
$\therefore \text{W}={\int }_{{\left(\text{V}\right)}_{\text{i}}}^{{\left(\text{V}\right)}_f}\left(\text{nRT}\right)\frac{\text{RV}}{\text{V}}$
$=\text{nRT}{\left|\text{ln V}\right|}_{{\text{V}}_{\text{i}}}^{{\text{V}}_f}=\text{nRT}ln\frac{{\text{V}}_f}{{\text{V}}_{\text{i}}}$
${\text{W}}_{\text{ab}}={\text{nRT}}_{\text{b}}ln\frac{2{\text{V}}_0}{{\text{V}}_0}=2\text{R}\times 500ln2=1000Rln2$
and ${\text{W}}_{\text{cd}}={\text{nRT}}_{\text{c}}ln\frac{{\text{V}}_0}{2{\text{V}}_0}=2\text{R}\times 300ln\frac{1}{2}=-600Rln2$
There is no volume change from b to c and from d to a, so
${\text{W}}_{\text{bc}}={\text{W}}_{\text{da}}=0$
The work done in the complete cycle
$\text{W}={\text{W}}_{\text{ab}}+{\text{W}}_{\text{bc}}+{\text{W}}_{\text{cd}}+{\text{W}}_{\text{da}}$
$=1000Rln2+0-600Rln2+0$
$=400Rln2$