Quasi-static, adiabatic problem

[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.

 

[Chestermiller]

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

 

[vela]

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.

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

 

[mateomy]

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

where c = 3/2

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

Chet

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...

 

[haruspex]

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

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

You exponentiated the LHS wrongly. e^ab is not e^ae^b.

 

[mateomy]

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

 

[Chestermiller]

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

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

Chet

 

[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.

 

[haruspex]

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

According to http://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_.28reversible_process.29 you have arrived at the correct equation. What answer do you get and what is the given answer?

 

[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.

 

[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?

 

[haruspex]

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.

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.

 

[Chestermiller]

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.

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

 

[mateomy]

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