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The figure given below shows the variation in the internal energy $U$ with volume $V$ of $2.0\ mole$ of an ideal gas in a cyclic process $abcda$ . The temperatures of the gas during the processes $ab$ and $cd$ are $500\ K$ and $300\ K$ respectively, the heat absorbed by the gas during the complete process is .... $J$
(Take $R$ = $8.3\ J/mol-K$ and $ln\ 2$ = $0.69$)
$3200$
$0$
$2100$
$2291$
Solution
Given an ideal gas whose $n=2.0$ moles
In the cyclic proces $\Delta \mathrm{U}=0$
Here $\mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} .$ As $\Delta \mathrm{U}=0$
the amount of heat absorbed.
$\mathrm{Q}=\mathrm{W}=\mathrm{W}_{\mathrm{ab}}+\mathrm{W}_{\mathrm{cd}}$
$=\mu \mathrm{RT}_{1} \ell n \left(\frac{2 \mathrm{V}_{0}}{\mathrm{V}_{0}}\right)+\mu \mathrm{RT}_{2} \ln \left(\frac{\mathrm{V}_{0}}{2 \mathrm{V}_{0}}\right)$
$=2 \times 8.3 \times 0.69(500-300)=2291 \mathrm{J}$
