Thermodynamics help.

1. Apr 22, 2006

IrAlien

I am stuck on a few questions in the thermodyanmics section. It is a revision sheet for the upcoming test. I have done about 2/3 of it but am stuck on a few questions. I will type up the questions in hopes of getting some help. Thank you in advance.

1. Derive a Maxwell relation by using the equality of the mixed second partial derivatives of the enthalpy H(S,p)

2. What thermodynamics state function does X represent in the following: dX = (p/nR)dV + (V/nR)dp

3. Beginning with dS = (1/T)dU + (p/T)dV and the definiton of the heat capacity at a constant volume, derive an expression for the entropy of an ideal monatomic gas as a function of the temperature T and volume V.

4. Calculate the change in entropy of an ideal gas when it undergoes an adiabatic free expansion starting from a state volume of 200cm^3, pressure 50kPa and temperature 300K to a final volume of 400cm^3.

And the last one,

5. An engineer announces that he has made a new type of heat engine with a greater theoretical maximum efficiency than the Carnot cycle. The new cycle consists of:
1. Adiabatic expansion from V1 to V2
2. Isobaric compression to V3
4. Isobaric expansion back to V1.
Derive an expression for the total work done by the engine in one cycle in terms of temperature and an expression for the heat absorbed in the isobaric expansion in terms of temperature.

Thanks again for the people who attempted to help or looked at this thread.

2. Apr 22, 2006

IrAlien

I worked out qn 3. So scrap that. :)

3. Apr 22, 2006

Andrew Mason

Begin by writing out the enthalpy function: H = U + PV

Substitute:

$$T = (\frac{\partial H}{\partial S})_P$$ and

$$V =(\frac{\partial H}{\partial P})_S$$

before taking the second partial derivative.

Multiply both sides by nR:

$$nRdX = PdV + VdP = d(PV)$$

That should tell you right away what dX is (think Ideal Gas law).

AM

Last edited: Apr 22, 2006
4. Apr 22, 2006

Andrew Mason

This is a little tricky.

Since the expansion is free, there is no external work done by the gas. The work is done to the gas itself, however, in expanding. So there is work done. Since the work is done to the gas and no heat is added or lost, we know that the total energy of the gas does not change, so the temperature does not change.

So: dU = dQ - PdV = = TdS - PdV = 0

We conclude that TdS = PdV or dS = PdV/T = nRdV/V (using PV=nRT)

The change in entropy of the gas is $\int ds$.

$$\Delta S = \int_{V_i}^{V_f} ds = \int_{V_i}^{V_f} \frac{nR}{V}dV = nRln(\frac{V_f}{V_i}) = nRln(2)$$

AM

Last edited: Apr 22, 2006
5. Apr 22, 2006

IrAlien

Thank you! I've done the Maxwell relation last night and got exactly what you got, but failed to get a working forumlae for 4 (Because I don't know what to put for "n" in dS = n.R.ln(v2/v1). n = pV/RT, R's cancel out but what value of "V" do I use? I'm still a little stuck on 5. How do I derive the work for an adiabatic expansion in that cycle? I look around but they don't give decent derivations. And since my professor likes to see derivations... -shudders-

Again, thank you Andrew Mason.

6. Apr 22, 2006

IrAlien

...Holy, I didn't combine the product of the derivatives. I guess now it's fairly obvious, dX is T. Lol...Thanks a lot again.

7. Apr 22, 2006

Andrew Mason

nR = PV/T where P is the pressure at a particular volume V and temperature T. So if P = 50 kPa and T = 300 K when V = 200 cm^3, what is nR?

$$PV^\gamma = \text{Constant} = K$$

From that you can work out the expression for work $\int Pdv$ by substituting: $P = KV^{-\gamma}$

$$\int_{V_i}^{V_f} PdV = K\int_{V_i}^{V_f}V^{-\gamma}dV$$

That gives you the work for adiabatic expansion/compression. The work integral under the isobars is just $P\Delta V$. Be sure to get the right signs and add them up

AM

8. Apr 22, 2006

IrAlien

Andrew,

I got to w = K( (V(final)^1-n) - V(initial)^1-n) )/ (1-n)
n = gamma

The question wants me to convert them all in terms of temperature. Can you give me a hint on where to start on changing that to Temperatures?

Also, anyone know how to start this question?

Use S = nR.ln((V.T^3/2)/n) + (3/2)nR((5/3) +C) to obtain an expression for the Helmholtz Free Energy of an ideal gas.

Last edited: Apr 22, 2006
9. Apr 23, 2006

Andrew Mason

Use: T = PV/nR. So T1 = P1V1/nR; $PV^\gamma = K$ so $T2 = KV_2^{1-\gamma}/nR$ ....etc.

AM

Last edited: Apr 23, 2006