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.

 

[DeeNos]

I would recommend first understanding the concept of adiabatic expansion and how it relates to pressure and temperature changes in a gas. Adiabatic expansion refers to a process in which there is no heat transfer between the system (the gas in this case) and its surroundings. This means that the change in temperature is solely due to the work done by or on the gas.

In this case, the gas is expanding against an external pressure of 1.162x10^5 Pa, which means that work is being done on the gas. This means that the gas is losing internal energy, which results in a decrease in temperature. This can be described by the adiabatic expansion equation:

P(initial)V(initial)^(gamma) = P(final)V(final)^(gamma)

Where gamma is the ratio of specific heats (Cp/Cv) and is equal to 5/3 for a monatomic gas.

Using this equation, you can solve for the final volume of the gas and then use the ideal gas law to calculate the final pressure. This will give you the final pressure after the stopper is replaced, which is equal to the external pressure.

To find the initial temperature, you can use the fact that the final temperature is equal to the initial temperature when the stopper is replaced. This can be described by the equation:

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

Since Cv,m = 3R/2 for a monatomic gas, you can substitute this in and solve for T(initial) using the final pressure calculated earlier.

I hope this helps in understanding the process and finding the final pressure for a monatomic gas. Good luck with your homework!