Adiabatic expansion against constant pressure

[MexChemE]

Homework Statement

a) For a certain ideal gas C_P = 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)

Homework Equations

P_1 V_1^\gamma = P_2 V_2^\gamma

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 C_P = 8.58 cal mol^-1 K^-1 and C_V = 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 C_P 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!

 

[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.

 

[Chestermiller]

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

 

[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!

 

[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 nC_VΔT = -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.

 

[Chestermiller]

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 nC_VΔT = -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.

Nicely done.

Chet

 

[MexChemE]

Thanks!