Question

Calculate the maximum work when 24 g of oxygen are expanded isothermally and reversibly from a pressure of $1.6\times {10}^5$ Pa to 100 kPa at 298 K.

Answer 1

Solution:-

Given weight $=24g$

Molecular weight $=32g$

No. of moles $\left(n\right)=\frac{24}{32}=0.75\text{mol}$

As we know that work done in an isothermal process is given as-

$W=-2.303nRT\log \frac{P_1}{P_2}$

Given:-

$T=298K$

$P_1=1.6\times {10}^{5}Pa=160kPa$

$P_2=100kPa$

Therefore,

$W=-2.303\times 0.75\times 0.0821\times 298\times \log \frac{160}{100}$

$\Rightarrow W=-42.26\times \log 1.6=-8.62atm-L$