Calculating Work in Isothermal Gas Expansion with Variable R_v

[alexialight]

Calculate the work W done by the gas during the isothermal expansion (path C). It may be be convenient to generalize your results by using the variable R_v, which is the ratio of final to initial volumes (equal to 4 for the expansions shown in the figure.)
Express W in terms of p_0, V_0, and R_v.

I've been at this question for ages and I just can't see how p_0 fits into the answer. The hints say to find an expression for p(V) in terms of p_0, V_0 and V and I can't even seem to do that. Any help would be much appreciated.

 

[dextercioby]

U can find it writing Mendeleev-Clapeyron's equation for the initial & final states.

U have to integrate it to get the answer.

Daniel.

 

[Corneo]

Knowing that

W = \int_{V_1}^{V_2} p ~dV

and

p = \frac {nRT}{V}

you can get

W=nRT \ln \frac {V_2}{V_1}

Since it is an isoterm, then T is constant.

p_1V_1 = p_2 V_2 or

\frac {V_2}{V_1} = R_V = \frac {p_1}{p_2}

This is a MP problem isn't it?

 

[alexialight]

Yes it is :(
I understand that W is equivilent to that equation you wrote, I just don't know how to get an answer with only p_0, V_0 and R_v and not n, T and R

 

[Corneo]

Well If I recall correctly, just because MP says you have to use these variables, it doesn't mean all the variables have to be used. But you can't use variables that aren't defined for the problem.

So you agree that

W=nRT \ln R_v

Since n, R, T are all constants in this problem.

p_0V_0 = nRT = pV, then perhaps your final answer is

W=nRT \ln R_v = p_0 V_0 \ln R_v

Enter that on your own risk, I never liked MP. Good luck.

 

[janiexo]

I had the same problem. It works now though :)