Calculate work and heat in a cyclic process ?

[rash219]

Homework Statement

Consider n moles of ideal gas kept in a cylinder with a piston. Two heat reservoirs 1 and 2 with the temperatures T1 < T2 are available, and at any given moment of time the heat exchange is established with only one of the reservoirs. In the initial equilibrium state the external pressure is p1, and heat contact is with reservoir 1. At the same time moment the external pressure is quickly changed to p2, whereas heat exchange switches to reservoir 2, and we wait for the system to equilibrate. Then the external pressure is quickly returned to its initial value p1, whereas heat exchange switches back to reservoir 1, and we wait for the system to equilibrate.

Calculate the amount of work (w) the system produce on the environment, the amount of heat (q2) transferred from reservoir 2 (heat source) to the system, and the amount of heat (q1) transferred to reservoir 1 from the system.

Homework Equations

dU = 0
U = q - w
w = -∫ Pext dV

The Attempt at a Solution

The system returns to its initial state and therefore we can call our entire process a cycle.

To calculate work total

Wtotal = w1 + w2

w = -∫ Pext dV

w1 = p2( v2 - v1)
w2 = - p1 (v1 -v2)

∴ Wtotal = p1v2 + p2v2 - p1v1 - p2v1

we do not know if this is correct and we have no idea how to proceed to calculate q1 and q2

 

[rude man]

Homework Statement

w1 = p2( v2 - v1)
w2 = - p1 (v1 -v2)

You're saying w1 and w2 both have the same sign which you know is not right.

Actually, I have problems with this problem - it doesn't seem to be a reversible process since we assume instantaneous state changes. But, assuming reversibility anyway:

How about w_1→2 = p2(V2-V1)
w_2→1 = -p1(V2-V1)

Then sum these two, substitute V = nRT/p as appropriate, and come up with net work done per cycle as a function of p1, p2, T1, T2, n and R.

Of course, you know that net work must equal Q2 -Q1. What second equation can you produce to enable solving for Q1 and Q2 separately? (Hint: what does the 2nd law require?)

Note: hopefully others will join in here to maybe give you a second opinion ...