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
$\text{W} = \displaystyle \int _{\text{V}_{\text{i}}}^{\text{V}_{f }} \text{PdV}$
We have,
$\text{PV} = \text{nRT} \Rightarrow \text{P} = \frac{\text{nRT}}{\text{V}}$
$\therefore \text{W} = \displaystyle \int _{\left(\text{V}\right)_{\text{i}}}^{\left(\text{V}\right)_{f }} \left(\text{nRT}\right) \frac{\text{RV}}{\text{V}}$
$= \text{nRT} \left|\text{ln V}\right|_{\text{V}_{\text{i}}}^{\text{V}_{f }} = \text{nRT} ln \frac{\text{V}_{f ⁡}}{\text{V}_{\text{i}}}$
$\text{W}_{\text{ab}} = \text{nRT}_{\text{b}} ln \frac{2 \text{V}_{0}}{\text{V}_{0}} = 2 \text{R} \times 5 0 0 ln 2 = 1 0 0 0 R ln 2$
and $\text{W}_{\text{cd}} = \text{nRT}_{\text{c}} ln \frac{\text{V}_{0}}{2 \text{V}_{0}} = 2 \text{R} \times 3 0 0 ln \frac{1}{2} = - 6 0 0 R ln 2$
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
$\text{W} = \text{W}_{\text{ab}} + \text{W}_{\text{bc}} + \text{W}_{\text{cd}} + \text{W}_{\text{da}}$
$=1000R ln 2+0-600R ln 2+0$
$= 4 0 0 R ln 2$