Work done on thermal systems confusion

[Ichiru]

There is this seemingly simple problem I have but dispite everything I've tried to do I can't find a way to remove one of the unknown variables. Everytime I try to remove it with some other equation I end up figuring out that I don't know another one of the values and end up not being able to solve it. Then there is other information I really am not sure how it would help figure out the problem.

Original Problem:

An external force compresses 0.10 mol of an ideal gas in thermal isolation. The curve of this transformation on a p-v diagram is pV^1.4 = a constant. The gas initially has a volume of 1.6L and a tempature of 275K. When the compression is finished, the tempature has increased to 380K. (a) What are the final volume and final pressure? (b)How much work was done on the gas to compress it?


My attempt at the solution:

When I initially tried the problem I immediatly went to the section in my book that it was on and looked for any equations that would help me.

The only equation I found was that Work = integral of pdV from V1 to V2

My brain said, hey I can use that to find the work after I find the values!

So then I tried looking for problems to help me find what the final volume and pressure were so I could solve part b.

I ended up using the equation p = nRT/V to find the inital volume which I found to be 1.43atm. After that I tried putting the equation for p into the integral equation to try and solve it but realized I didn't know what work was so I couldn't solve it that way.

After that I tried searching my textbook for clues but as of yet found nothing.

Thank you for your time and any and all help is greatly appreciated

 

[nicktacik]

For isothermal expansion you have

PV^{\gamma}=C

Which you can re-write as

P(V)V^{\gamma} = P_1 V_{1}^{\gamma}

Solving for P(V) you get

P(V) = P_1 V_{1}^{\gamma}V^{-\gamma}

Which makes the work integral

W = P_1 V_{1}^{\gamma}\int_{V1}^{V2}V^{-\gamma}dV

 

[Ichiru]

Thanks! I think I understand how to use that.