Entropy Change For Van Der Waals Gas

[laser1]

Homework Statement

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Relevant Equations

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I have two different methods giving different results

Why is this the case? (Left method was answer in solutions, right method was my answer before checking the solutions). Also yes pretend V was V_1 or something ignore my dummy variables :)

 

[Chestermiller]

The first method is correct. Why do you feel the 2nd method is correct?

 

[laser1]

The first method is correct. Why do you feel the 2nd method is correct?

I don't see any flaws in logic. I simply substitute for P and integrate.

 

[Chestermiller]

I don't see any flaws in logic. I simply substitute for P and integrate.

Your equation for the effect of pressure on entropy in the first cast is correct. Where did you get the starting equation for the effect of pressure on entropy in the 2nd case from?

 

[laser1]

Your equation for the effect of pressure on entropy in the first cast is correct. Where did you get the starting equation for the effect of pressure on entropy in the 2nd case from?

The question states "isothermal", which implies $U=0$. So by the first law, this must mean that $dQ=PdV$. Following this, the definition of the change in entropy is $$\Delta S = \int_i^f{\frac{dQ}{T}}$$ and I substitute $dQ=PdV$ to get the starting equation for #2.

 

[Chestermiller]

The question states "isothermal", which implies $U=0$. So by the first law, this must mean that $dQ=PdV$. Following this, the definition of the change in entropy is $$\Delta S = \int_i^f{\frac{dQ}{T}}$$ and I substitute $dQ=PdV$ to get the starting equation for #

Isothermal does not imply that du is equal to 0 for a Vdw gas.