Work done in Adiabatic, Quasistatic-compression

[Trenthan]

Homework Statement

i have to find the W done in an Adiabatic, Quasistatic compression and I am not having much luck.

Ive calculated it using (delta)U = Q + W, since Q = 0, hence W = change in internal energy

However i want to calculate it using W = -(delta)P*V

The Attempt at a Solution

My working is attached, my text just says to use W = delta(U), but i know there is a way to do it via pressure and volume, I am just not sure where to go from where I am at.

Attempt is attached

Has to be in terms of T_A, n (moles) and R

Cheers Trent

 

[BobbyBear]

I assume we are talking about perfect gases so that indeed p V^\gamma = const for a quasistatic, adiabatic process :p

<br /> <br /> -\int_{V_C}^{V_A} p dV = <br /> <br /> \left\{ p V^\gamma = const \right\}= <br /> <br /> -\int_{V_C}^{V_A} const \cdot \frac{dV}{V^\gamma} <br /> <br />

Determine const from the conditions at A.

 

[Trenthan]

I assume we are talking about perfect gases so that indeed p V^\gamma = const for a quasistatic, adiabatic process :p

<br /> <br /> -\int_{V_C}^{V_A} p dV = <br /> <br /> \left\{ p V^\gamma = const \right\}= <br /> <br /> -\int_{V_C}^{V_A} const \cdot \frac{dV}{V^\gamma} <br /> <br />

Determine const from the conditions at A.

I got it finally thanks heeps but i got it using conditions at "c" our starting point, (P_CV_C^gamma = PV^gamma, rearranged for P = and solved from ther) not "A", I am assuming u meant "c"?

Thanks :)

 

[BobbyBear]

Oopsie, yes I meant "C" , it's the point where you know P and V to start with. Sorry, I always think that A comes before C when I'm not thinking :P:P

Though now that you said, . . . it's also true that you know P and V of point A too, so you could use any:) of course, they've given you two points such that :
<br /> <br /> <br /> p V^\gamma = const1<br /> <br /> <br />
and
<br /> <br /> <br /> T V^{\gamma-1}= const2<br /> <br /> <br />
with
<br /> <br /> <br /> \gamma=5/2<br /> <br /> <br />
because if this was not so it wouldn't be a reversible, adiabatic process.

 

[Trenthan]

thanks and good point, just trierd it the other way and it works :)