Thermodynamics, Callen's book problem

[fluidistic]

Homework Statement

The first problem in the book reads:
A given system is such that quasi-static adiabatic change in volume at constant mole numbers is found to change the pressure in accordance with the equation $P= KV^{-5/3}$ where K is a constant.
Find the quasi-static work done on the system and the net het flux to the system in each of the 3 processes indicated in the figure. Each process is initiated in the state A, with a pressure of 32 atm and a volume of 1 liter, and each process terminates in the state B, with a pressure of 1 atm and a volume of 8 liters.
Process a: The system is expanded from its original to its final volume, heat being addd to maintain the pressure constant. The volume is then kept constant and heat is extrated to reduce the pressure to 1 atm.

(I might describe the other processes, only if I have troubles in doing them later)

Homework Equations

$dW=-PdV$
$dQ=dU+PdV$
Answer is W=-224 liter-atm and Q=188 liter-atm.

The Attempt at a Solution

I tried to use the fact that $P= KV^{-5/3}$, integrated dW from A to D. This gave me $\frac{3}{2}K(8^{-2/3}-1)$. So I'd get an answer depending on K which isn't acceptable apparently.
Another try was to write $W=-\int _A ^D PdV-\int _D ^B PdV=-P(V_D-V_A)-\int _D ^B PdV$ but in the last integral P decreases at a constant volume so it's not a function of V. I don't know how to perform such an integration.

P.S.:I don't think the figure is required to understand the situation. It is what the problem statement is.
Thanks for any help.

 

[rude man]

The Attempt at a Solution

I tried to use the fact that $P= KV^{-5/3}$, integrated dW from A to D. This gave me $\frac{3}{2}K(8^{-2/3}-1)$. So I'd get an answer depending on K which isn't acceptable apparently. No, because neither sub-process is adiabatic!

Another try was to write $W=-\int _A ^D PdV-\int _D ^B PdV=-P(V_D-V_A)-\int _D ^B PdV$ but in the last integral P decreases at a constant volume so it's not a function of V. I don't know how to perform such an integration.

.

What are your states C and D? There is only one intermediate state. I'll call it I.

Now then: did you draw a p-V graph of the two sub-processes? If so, the work W done from A to I and from I to B is immediately seen!

As for Q: here you need to look carefully at the given p-V relationship under an adiabatic process. What does it tell you about the kind of gas you're given?

The hard way is to work this backwards but a far easier way is to remember what you learned in class so far ...

 

[fluidistic]

What are your states C and D? There is only one intermediate state. I'll call it I.

Now then: did you draw a p-V graph of the two sub-processes? If so, the work W done from A to I and from I to B is immediately seen!

Thanks, I get it. To answer my questions in my last post, the last integral is worth 0 (no area under the curve) and the first integral is easily computed as the pressure is constant. So $W=-P \delta V$.

As for Q: here you need to look carefully at the given p-V relationship under an adiabatic process. What does it tell you about the kind of gas you're given?

The hard way is to work this backwards but a far easier way is to remember what you learned in class so far ...

Here I am not sure. We've just started classes last week and learned nothing about ideal gases (well I learned about that in my first year in an introductory course, but currently not in the advanced undergrad course). Does that mean that the gas is ideal?

 

[rude man]

Good guess, but you need to find out even more than that. Hint: there are monatomic gases and there are diatomic gases and there are ...

Reason you need to figure it out is certain heat capacities you will need presently.

The, re-write the 1st law in differential form, replace dU and dQ with the appropriate terms invoking heat capacities. You're doing this one process at a time, so this is from A to I.

 

[paranormal]

As a scientist, my response would be to approach the problem by first identifying the relevant equations and principles from thermodynamics. In this case, we can use the first law of thermodynamics: dQ = dU + PdV, and the definition of work: dW = -PdV.

To solve for the work done on the system, we can integrate dW from the initial state A to the final state B. Using the equation P= KV^{-5/3}, we can substitute for P and solve the integral:

W = -∫dW = -∫PdV = -∫KV^{-5/3}dV

Integrating from V=1 liter to V=8 liters, we get:

W = -K∫V^{-5/3}dV = -K[3V^{-2/3}]_1^8 = -3K(8^{-2/3}-1) = -3K(1/4-1) = -3K(-3/4) = 9/4K

Since we are given that the pressure is constant during the first part of the process (from A to D), we can use the equation dQ = dU + PdV to solve for the heat added to the system during this part:

dQ = dU + PdV = 0 + PdV = KV^{-5/3}dV

Integrating from V=1 liter to V=8 liters, we get:

Q = ∫dQ = ∫KV^{-5/3}dV = K[3V^{-2/3}]_1^8 = K(3/4-1/4) = K(1/2)

Finally, for the second part of the process (from D to B), the volume is kept constant while the pressure decreases. This means that no work is done (dW=0) and the heat extracted can be calculated using the same equation as before:

dQ = dU + PdV = 0 + PdV = KV^{-5/3}dV

Integrating from V=8 liters to V=1 liter, we get:

Q = ∫dQ = ∫KV^{-5/3}dV = K[3V^{-2/3}]_8^1 = K(1/4-8/4