To get an expansion of Helmholtz energy of

a) an ideal gas b) a Van der waals gas

we must integrate

$\left ( \frac{\delta A }{\delta V} \right )_{T}=-P$

I saw the solution is :

enter image description here

Can you explain why is that?


Consider just part (a). we use the ideal gas law for $P$, that is $$-P=-\frac{nRT}{V}$$ and substituting this in for $P$ we get $$\delta A=-\frac{nRT}{V}\delta V\implies\Delta A=-nRT\int_{V_1}^{V_{2}}\frac{1}{V}dV$$ which gives $$\Delta A=-nRT(\ln(V_2)-\ln(V_1))=-nRT\ln\left(\frac{V_2}{V_2}\right)$$ I'm not sure where the $n$ went... It's key we held $T$ constant. Hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.