Question

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

Answer 1

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 $=-P_{ext }ΔV$
$=-4.3\times \left(\frac{5 \times 293 R}{1 . 3} - \frac{5 \times 293}{2 . 1}\right)$
$=-5\times 293\times 8.314\times 43\left(\frac{1}{13} - \frac{1}{21}\right)$
$=\frac{5 \times 293 \times 8 . 314 \times 43 \times 8}{21 \times 13}$
$=-15347.7049J$
$=-15.34KJ$
Isothermal process, so $ΔU=0$
$w=-Q$
$Q=15.34KJ/mol$
So answer is $15$