1. Nov 1, 2013

### mateomy

Two moles of a monatomic ideal gas are at a temperature of 0°C and a
(dQ = 0) and quasi-static volume of 45 liters. The gas is expanded adiabatically
until its temperature falls to - 50°C. What are its initial and final pressures and its final volume?

I've been beating my head against a wall for a while now on this one.

I know how to find the initial pressure. That's just an easy PV=nRT problem. I get an answer of 100,956 Pa (0.1MPa). I can't find the final pressures though, totally stuck.

So what I've done thus far is to utilize the first law in that I know dQ = 0. So I know that dU = -dW. Solving the U = cNRT for initial and final temperatures and taking the difference, I get an energy which I know must be attributed to the work done. This is where I get stuck.

Not sure what steps to take next. Need some assistance, thanks.

I thought I could try
$$W = - \int PdV$$
which I could do some substitution to get to (eventually),
$$W = -nRTln\left(\frac{V_f}{V_i}\right)$$
But when I do that and I set W to the difference in energy I found earlier, I get an answer that doesn't match my book's.

2. Nov 1, 2013

### Staff: Mentor

No, but you almost have it. Going back to your equation for the first law:

dU=cNRdT=-PdV=-(NRT/V)dV

where c = 3/2

Cancel NR from both sides, divide both sides by T, and integrate.

Chet

3. Nov 1, 2013

### vela

Staff Emeritus
This doesn't work because that expression for the work done is for isothermal processes. That's not what you have here.

4. Nov 1, 2013

### mateomy

So this is what I've done on your suggestion (asking for my work to be checked because my answer still isn't correct):

$$cNRdT = -\frac{NRTdV}{V}$$

$$\frac{cdT}{T} = \frac{dV}{V}$$

$$c \int_{T_i}^{T_f} \frac{dT}{T} = - \int_{V_i}^{V_f} \frac{dV}{V}$$

$$c\,ln\left(\frac{T_f}{T_i}\right) = - ln\left(\frac{V_f}{V_i}\right)$$

$$e^{c} (T_f - T_i) = -(V_f - V_i)\,\,\,\,\,\,negatives\,cancel\,here$$

from which I get a result of 224.04. However, the answer is $61E3$. Still unsure...

5. Nov 1, 2013

### haruspex

You exponentiated the LHS wrongly. eab is not eaeb.

6. Nov 1, 2013

### mateomy

I'm insanely confused then. Is it,
$$\left(\frac{T_f}{T_i}\right)^{c}$$

7. Nov 1, 2013

### Staff: Mentor

Now you're cookin'. Of course, you follow the same recipe on the RHS of the equation.

Chet

8. Nov 1, 2013

### mateomy

Now I'm becoming embarrassed...
$$\left(\frac{T_f}{T_i}\right)^{c} = \left(\frac{V_f}{V_i}\right)^{-1}$$
$$\left(\frac{T_f}{T_i}\right)^{c} = \left(\frac{V_i}{V_f}\right)$$
$$V_i \left(\frac{T_i}{T_f}\right)^{c} = V_f$$
But I'm still not getting the correct answer.

9. Nov 1, 2013

### haruspex

10. Nov 1, 2013

### mateomy

The given answer (from Callen) is $61 x 10^{3} m^{3}$. My answer is 0.06095. It almost seems like there's a conversion error but I don't see it. From the initial values: 45 liters = 0.045 cubic meters, 0 C = 273K, and -50 C = 223K.

11. Nov 1, 2013

### mateomy

Although...

I'm looking at my PDF (don't have the 'real' book) and it's entirely possible that the negative is smudged out. I've noticed a lot of cosmetic errors in this PDF. That seems reasonable if my equation is correct and I am off by that magnitude. Maybe someone could check my numbers to confirm?

12. Nov 1, 2013

### haruspex

No need to check the calculation in detail. It's quite clear that the volume would not expand by more than a factor of 10, probably much less. So 61 litres is reasonable.

13. Nov 1, 2013

### Staff: Mentor

Yeah. That should be 10-3. You did a really nice job on this problem. Too bad the book had that typo.

14. Nov 1, 2013

### mateomy

Ugh. I can't believe it was a typo. At least I got some lessons out of it. Thank you, both.