Question

Five moles of an ideal gas at $293K$ is expanded isothermally from an initial pressure of $2.1MPa$ to $1.3MPa$ against at constant external $4.3MPa$ . The heat transferred in this process in $kJmo{l}^{-1}$ is (Rounded-off of the nearest integer) [Use $R=8.314Jmo{l}^{-1}{K}^{-1}$ ]

Answer 1

Answer: 15

Moles $\left(n\right)=5$
$T=293k$
It is irreversible isothermal expansion.
$P_{ini}=2.1$ $MPa$
$P_t=1.3MPa$
$P_{ext}=4.3mPa$
Work $=-P_{ext}\Delta V$
$=-4.3\times \left(\frac{5\times 293R}{1.3}-\frac{5\times 293}{2.1}\right)$
$=-5\times 293\times 8.314\times 43\left(\frac{1}{13}-\frac{1}{21}\right)$
$=\frac{5\times 293\times 8.314\times 43\times 8}{21\times 13}$
$=-15347.7049J$
$=-15.34KJ$
Isothermal process, so $\Delta U=0$
$w=-Q$
$Q=15.34KJ/mol$
So answer is $15$