Question

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

Answer 1

Given $n=2$ moles

$dV=0$

in $ad$ and $bc$

Hence $dW=dQ$ $dW=d{W}_{ab}+d{W}_{cd}$

$=nR{T}_{1}Ln\frac{2V_0}{V_0}+nR{T}_{2}Ln\frac{V_0}{2V_0}$

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

$=2\times 8.314\times 2.303\times 0.301\times 200=2305.31J$