Work done in Adiabatic, Quasistatic-compression

[Trenthan]

Homework Statement

i have to find the W done in an Adiabatic, Quasistatic compression and im 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, im just not sure where to go from where im 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

$$-\int_{V_C}^{V_A} p dV = \left\{ p V^\gamma = const \right\}= -\int_{V_C}^{V_A} const \cdot \frac{dV}{V^\gamma}$$

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

$$-\int_{V_C}^{V_A} p dV = \left\{ p V^\gamma = const \right\}= -\int_{V_C}^{V_A} const \cdot \frac{dV}{V^\gamma}$$

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", im 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 :
$$p V^\gamma = const1$$
and
$$T V^{\gamma-1}= const2$$
with
$$\gamma=5/2$$
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 :)