Question

Figure shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a 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.

Answer 1

Given, n = 2 moles.

$\Delta \text{V}=0$

In ad and bc

Hence $\Delta \text{W}=\Delta {\text{W}}_{\text{AB}}+\Delta {\text{W}}_{\text{CD}}$

$={\text{nRT}}_{1}ln\left(\frac{2{\text{V}}_0}{V_0}\right)+{\text{nRT}}_{2}ln\left(\frac{{\text{V}}_0}{2V_0}\right)$

$=\text{nR}\times 2.303\times \text{log 2}\times (500-300)$

$=2\times 8.314\times 2.303\times 0.301\times 200$

$=2305.31\text{J}$