# Adiabatic expansion against constant pressure

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1. Nov 3, 2014

### MexChemE

1. The problem statement, all variables and given/known data
a) For a certain ideal gas CP = 8.58 cal mol-1 K-1. We have 2 moles of said gas at 293.15 K and 15 atm. Calculate the final volume and temperature when the gas expands adiabatically and reversibly until it reaches a pressure of 5 atm. (Answers: V = 7.45 L; T = 227.15 K)
b) Consider the same gas, but now the expansion takes place adiabatically and reversibly against a constant pressure of 5 atm. Calculate the final pressure and volume of the gas. (Answers: V = 8.15 L; T = 248.15 K)

2. Relevant equations
$$P_1 V_1^\gamma = P_2 V_2^\gamma$$

3. The attempt at a solution
For a)
I can calculate V1 via the ideal gas law, which gives V1 = 3.207 L. Now I have P1, V1 and P2, so I can clear V2 from the adiabatic process equation. I have CP = 8.58 cal mol-1 K-1 and CV = 6.593 cal mol-1 K-1. Thus, γ = 1.3.
$$V_2 = (3.207 \ L) \left(\frac{15 \ atm}{5 \ atm} \right)^{\frac{1}{\gamma}} = 7.466 \ L$$
Now I can get T2 via the ideal gas law, which gives T2 = 227.48 K.

For b)
I have no idea how to tackle this part. I know CP and n remain the same, but I don't know what pressure to take as P1 and what to take as T1 to get to the correct answers provided by the textbook. Besides, if pressure is to remain constant, isn't the adiabatic process equation telling me the volume should remain constant too? As you can see, I'm a little bit lost on this one. Thanks in advance for any input!

2. Nov 3, 2014

### Feodalherren

An adiabatic, reversible process follows PV^γ=Constant.

γ=Cp/Cv
Cp=Cv+R

Your P1 and T1 are ALWAYS just wherever the process started. Draw a P-V diagram and it will be obvious.

3. Nov 4, 2014

### Staff: Mentor

I am in agreement with you regarding part b. I think what they meant to say was adiabatically and irreversibly. Try the problem that way and see what you get.

Chet

4. Nov 4, 2014

### MexChemE

Indeed, the problem meant irreversibly. It turned out to be an algebraic puzzle more than a thermo problem. I'm a little busy right now so I'll post my solution tomorrow. Thank you both!

5. Nov 5, 2014

### MexChemE

For part b)
The initial state of the gas will be the same as in part a). P1 = 15 atm, V1 = 3.207 L and T1 = 293.15 K. The final pressure will be again P2 = 5 atm, but this time the expansion will happen irreversibly against a constant opposing pressure, p = 5 atm. We can calculate V2 as a function of T2, this will prove useful in the following procedure.
$$V_2 = \frac{(2 \ mol)(0.08205 \ \frac{L \cdot atm}{mol \cdot K})}{5 \ atm} T_2 = \left(0.03282 \ \frac{L}{K} \right) T_2$$
We are having an adiabatic process, so ΔU = -W, which equals nCVΔT = -pΔV. We plug our values into this last equation, we originally have two unknowns, V2 and T2, but we use our expression for V2 as a function of T2 and now we just have to solve for one unknown.
$$(2 \ mol) \left(0.27221 \ \frac{L \cdot atm}{mol \cdot K} \right)(T_2 - 293.15 K) = -(5 \ atm) \left( \left(0.03282 \ \frac{L}{K} \right)T_2 - 3.207 \ L \right)$$
After solving that monster we get T2 = 247.89 K = -25.26 °C. Now with this temperature we can get the final volume, which gives V2 = 8.14 L.

6. Nov 5, 2014

Nicely done.

Chet

7. Nov 5, 2014

Thanks!