What Is the Final Pressure for a Monatomic Gas After Adiabatic Expansion?

[S.L.G.]

Homework Statement

A bottle at 323 K contains an ideal gas at a pressure of 1.667×10^5 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against P(ext)=1.162×10^5 Pa, and some gas is expelled from the bottle in the process. When P=P(ext), the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 323 K.

I have to find the final pressure for a monatomic gas...that is, Cv,m=3R/2.

Homework Equations

T(final) = T(initial) ((Cv,m + (RP(ext)/P(initial)))/(Cv,m + (RP(ext)/P(final)))

The Attempt at a Solution

I know that P(initial) = P(external) (making this a reversible process), but I don't know how to find T (initial) (after the stopper is removed)...

Thank you!

 

[JohnRC]

No way is this a reversible process! You are given p(initial) = 166.7 kPa and p(external) = 116.2 kPa for the adiabatic expansion.
For the second part of the process, warming in the sealed bottle, the initial pressure is equal to the external pressure, but the external pressure can play no part in a process inside a sealed bottle!
You are going to need to rethink your total approach to this problem, I fear.

 

[S.L.G.]

I know that this is in two steps.

Step 1: The bottle is initially closed; p(initial) = 166.7 kPa. Then the bottle is opened.

Step 2: p(external) = 116.2 kPa. When P=P(external), the bottle is closed again, at which point p(initial) = 116.2 kPa. I have to find p(final)...

 

[Chestermiller]

Homework Statement

A bottle at 323 K contains an ideal gas at a pressure of 1.667×10^5 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against P(ext)=1.162×10^5 Pa, and some gas is expelled from the bottle in the process. When P=P(ext), the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 323 K.

I have to find the final pressure for a monatomic gas...that is, Cv,m=3R/2.

Homework Equations

T(final) = T(initial) ((Cv,m + (RP(ext)/P(initial)))/(Cv,m + (RP(ext)/P(final)))

The Attempt at a Solution

I know that P(initial) = P(external) (making this a reversible process), but I don't know how to find T (initial) (after the stopper is removed)...

Thank you!

p_initial doesn't equal p_external. p_final equals p_external, where p_final refers to the pressure within the bottle immediately after the irreversible adiabatic expansion, but before the temperature has had a chance to rise to 323K. "When p = p_ext, the stopper is replaced." The equation you give tells you the temperature in the bottle immediately after the irreversible adiabatic expansion has taken place. Knowing the pressure in the bottle (p_external) and the temperature in the bottle (T_final) at the end of the irreversible adiabatic expansion gives you the information you need to calculate the number of moles per unit volume in the bottle at the end of the expansion. You can then use the ideal gas law to calculate the pressure when the temperature rises to 323K.