Why entropy change is different

[laser1]

Homework Statement

see description

Relevant Equations

$$\Delta S = \int_i^f{\frac{dQ_{rev}}{T}}$$

I tried two different methods when solving this question and have no idea why.

Method 1:
Using the Maxwell relation of $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V=\frac{R}{V-b}$$ then integrating it, I get $$\Delta S = \int_i^f\frac{R}{V-b}dV$$

Method 2:
$$\Delta S = \int_i^f\frac{P}{T}dV$$ as isothermal implies that $dQ=PdV$. Next, substituting in $P$ gives $$\int_i^f\left({\frac{R}{V-b}-\frac{a}{TV^2}}\right)dV$$

 

[Chestermiller]

Homework Statement: see description
Relevant Equations: $$\Delta S = \int_i^f{\frac{dQ_{rev}}{T}}$$

View attachment 352504
I tried two different methods when solving this question and have no idea why.

Method 1:
Using the Maxwell relation of $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V=\frac{R}{V-b}$$ then integrating it, I get $$\Delta S = \int_i^f\frac{R}{V-b}dV$$

Method 2:
$$\Delta S = \int_i^f\frac{P}{T}dV$$ as isothermal implies that $dQ=PdV$. Next, substituting in $P$ gives $$\int_i^f\left({\frac{R}{V-b}-\frac{a}{TV^2}}\right)dV$$

Method 2 is incorrect because your starting equation is incorrect.

 

[laser1]

Method 2 is incorrect because your starting equation is incorrect.

$dQ=PdV$ in isothermal process is incorrect? I am not sure why it is incorrect. Thanks

 

[Chestermiller]

$dQ=PdV$ in isothermal process is incorrect? I am not sure why it is incorrect. Thanks

Because, in an isothermal process for a real gas, dU is not zero.