Question

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

Answer 1

Number of moles of gases (n) = 5
$\text{Temperature},T=293\text{K}$
Given process is isothermal.
$P_{ini}=2.1\text{M Pa}$
$P_{final}=1.3\text{M Pa}$
$P_{ext}=4.3\text{MPa}$
Work $=-P_{\text{ext}}\Delta V$
Where, $\Delta V$ is change is volume.
$\therefore W=-4.3\times (\frac{5\times 293R}{1.3}-\frac{5\times 293R}{2.1})$
$=-5\times 293\times 8.314\times 43(\frac{1}{13}-\frac{1}{21})$
$=-\frac{5\times 293\times 8.314\times 43\times 8}{21\times 13}$
$=-15347.7049\text{J}$
$=-15.34\text{kJ}$
Isothernal process, so $\text{Internal energy}\left(\Delta U\right)=0$
$w=-Q$
$\therefore Q=15.34\text{kJ}$
So the heat transfered in this process is
$\frac{15.34}{5}{\text{kJmol}}^{-1}=3.068{\text{kJmol}}^{-1}$