Question

Isothermally at ${27}^{\circ }C$, one mole of a Van der Waal's gas expands reversibly from 2 liters to 20 liters. Calculate the work done if $a=1.42\times {10}^{12}dynesc{m}^{4}permole$ and $b=30cc$

• A. - 58.12 k J
• B. -861.2 k J
• C. -691.2 k J
• D. -761.2 k J

Answer 1

The correct option is A - 58.12 k J

Try to read the question properly.

Here the gas is vanderwaal's gas not ideal gas.

Do you recall the equation of vanderwaal's gas from last chapter

Let's try to recall it.

It was $P=\frac{RT}{V-b}-\frac{a}{V^2}$

Now, we have

$W=\underset{V_1}{\overset{V_2}{\int }}PdV=\underset{V_1}{\overset{V_2}{\int }}(\frac{RT}{V-b}-\frac{a}{V^2})dV$

= ${\left[RTln\right(V-b)+\frac{a}{V}]}_{V_1}^{V_2}=RTln\frac{V_2-b}{V_1-b}+a[\frac{1}{V_2}-\frac{1}{V_1}]$

= $2.303\times 8.31\times {10}^7\times 300log\frac{20-0.03}{2-0.03}+1.42\times {10}^{12}\left[\frac{2-20}{20\times 2}\right]$

= $-58124.67\times {10}^{7}ergs$

= $-58124.67Joules$