Callen Thermodynamics Problem

[Eldorin]

Homework Statement

A particular gas is enclosed in a cylinder with a moveable piston. It is observed that if the walls are adiabatic, a quasi-static increase in volume results in a decrease in pressure according to the equation
a ) Find the quasi-static work done on the system and the net heat transfer to the system in each of the three processes (ADB, ACB, and the direct linear process AB ) as shown in the figure.
In the process ADB the gas is heated at constant pressure ( P = 10^5 Pa) until its volume increases from its initial value of 10^-3 m^3 to its final value of 8e-3 m^3. The gas is then cooled at constant volume until its pressure decreases to 10^5/32 Pa. The other processes (ACB and AB) can be similarly interpreted, according to the figure.

P^3*V^5 =Constant for Q=0

I am currently only having trouble with the part with the direct linear process AB.

Homework Equations

dQ = δU - δW
W = -PdV

The Attempt at a Solution

When it says it is directly linear, I took that to mean the following:

P*V = Px*Vx
Where Px and Vx are the initial pressure and volume. From there I integrated the work equation and got the following:

W = -Px*Vx*ln(Vb/Va).

I then did the same thing with the P^3*V^5 = Constant equation, but when I used these, as worked with the previous parts. My answers did not match the ones given in the book (W = - 360.9 J and Q = 248.4J)

I appreciate any help, thanks!

 

[TSny]

Hello, Eldorin. Welcome to PF!

"Linear process" means that P is a linear function of V so that the graph is a straight line on the PV diagram.

So, that means P = aV + b for certain constants a and b. However, you shouldn't need to use this equation since you can determine the work W graphically from the PV diagram.

 

[Eldorin]

Still having a bit of trouble

Alright, so I tried going through this and experienced some additional problems.

So, I'd like to know how to solve it both graphically and not graphically.

I linearized the equation, P=a*V+b, where P1=10^5, V1 = 10^-3, P2 = 10^5/32, V2 = 8e-3.

From this I determined that a = -1.384, b=10^5 (Both rounded).

I then set up dU = dW = -PdV, integrated and found that Ub-Ua = 0.5*(1.384*(Vb^2-Va^2))-(10^5(Vb-Va)) = -700J.

From there, I graphically found (What I now assume to be Q, since it gets me the correct answer later in the progression. Can anyone explain why this is, but the Q answer I get doesn't match the answer?) by solving for the area under the triangle and got Q=339.06J. From there I set up

Ub-Ua = Q - W, and solved for W and got -360.94J.

At this point, I'm pretty thoroughly confused. Can anyone explain what I did wrong, how to find the solution without using the graph, and what solving it graphically actually gives, since it appears to not be Q or W.

Thanks much!

 

[TSny]

Your value of a is off by a certain power of 10. That makes your value of b incorrect.

When finding the work graphically, you need to find the area between the linear path and the V axis. This is the area of a trapezoid.

For the linear process, dQ is not zero.

If you already found ΔU for one of the other paths, does that help you find ΔU for the linear path? Remember, U is a state variable.

 

[Eldorin]

One more quick question so I can make sure I understand this completely. I've gotten the correct answer now, so it's no longer a problem with that, but more a conceptual problem.

Why are we allowed to assume that Q=0 for the P^3*V^5 = constant in order to simplify the equation from dU=dW+dQ to only dU=dW. Is this because we are assuming that it is a quasi-static work (i.e. done so slowly that none of the work is converted into waste heat) or is there another reason?

Thanks again for all of your help!

 

[TSny]

Why are we allowed to assume that Q=0 for the P^3*V^5 = constant in order to simplify the equation from dU=dW+dQ to only dU=dW. Is this because we are assuming that it is a quasi-static work (i.e. done so slowly that none of the work is converted into waste heat) or is there another reason?

It is not because the work is quasi-static that Q equals zero. For example, in all three processes in your problem, you would assume the work is quasi-static so that the pressure is well-defined at each point of the process. Quasi-static just means that the process is slow enough that the system is essentially in thermal equilibrium at each point of the process. Essentially, any time you see a continuous path drawn on a PV diagram, you can assume that the process is quasi-static.

The reason that Q = 0 for the process in which P³V⁵ = constant is because that is what was given to be true in the statement of the problem. (The word "adiabatic" implies Q = 0.)