Five moles of an ideal gas at $293 \mathrm{~K}$ is expanded isothermally from an initial pressure of $2.1 \mathrm{MPa}$ to 1.3 MPa against at constant external 4.3 MPa. The heat transferred in this process is__________ $\mathrm{kJ}$ $\mathrm{mol}^{-1}$. (Rounded-off of the nearest integer)
[Use $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ]
(15)
Moles $(n)=5$
$T=293 \mathrm{k}$
Process $=$ IsoT. $\rightarrow$ Irreversible
$P_{\mathrm{ini}}=2.1 \mathrm{M} \mathrm{Pa}$
$P_{t}=1.3 \mathrm{M} \mathrm{Pa}$
$P_{\mathrm{ext}}=4.3 \mathrm{mPa}$
Work $=-P_{\text {ext }} \Delta v$
$=-4.3 \times\left(\frac{5 \times 293 R}{1.3}-\frac{5 \times 293}{2.1}\right)$
$=-5 \times 293 \times 8.314 \times 43\left(\frac{1}{13}-\frac{1}{21}\right)$
$=\frac{5 \times 293 \times 8.314 \times 43 \times 8}{21 \times 13}$
$=-15347.7049 \mathrm{~J}=-15.34 \mathrm{KJ}$
Isothermal process, so $\Delta \mathrm{U}=0 \mathrm{w}=-\mathrm{Q}$
$\mathrm{Q}=15.34 \mathrm{KJ} / \mathrm{mol}$
So answer is 15
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