Calculate the work done by an external agent during an isothermal compression of$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{m}\mathit{o}\mathbf{\text{l}}$of oxygen from a volume of$\mathbf{22}\mathbf{.}\mathbf{4}\mathit{L}$at$\mathbf{0}\mathbf{°}\mathit{C}$and$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$to a volume of$\mathbf{16}\mathbf{.}\mathbf{8}\mathit{L}$.

• Number of moles$n=1.00\text{\hspace{0.17em}mol}$
• Temperature$T=0^{0}C=273K$
• Initial volume of gas$v_i=22.4\text{\hspace{0.17em}L}$
• Final volume of gas$v_f=16.8\text{\hspace{0.17em}L}$

In case of isothermal process temperature remains constant. The expression for the work done in case of isothermal process is given by,

$W=nRT\text{\hspace{0.17em}ln}\left(\frac{v_f}{v_i}\right)$

Here W is the work done,n is the number of moles, R is the gas constant,T is the temperature, $v_f$is the final volume of the gas, $v_i$is the initial volume of the gas.

The expression of the work done by an external agent is given by,

$W=nRT\text{\hspace{0.17em}ln}\left(\frac{v_f}{v_i}\right)$

Substitute$1.00\text{\hspace{0.17em}mol}$ for n, $8.31\frac{\text{J}}{\text{mol}}K$ for R, $273K$ for T , $16.8L$for$v_f$ , $22.4L$for $v_i$into the above equation,

$\begin{array}{c}W=\left(1.00\text{\hspace{0.17em}mol}\right)\left(8.31\frac{\text{J}}{\text{mol}}K\right)\left(273K\right)\text{ln}\left(\frac{16.8}{22.4}\right)\\ =-653\text{\hspace{0.17em}J}\end{array}$

Therefore, the work done by an external agent during the isothermal compression is$-653\text{J}$ .