*Thermodynamics*: Adding air to a bottle, finding internal energy and so on

[jolierouge]

Homework Statement

In class we saw a demo where a pop bottle was pumped up with extra
air. If we consider the bottle was pumped up to a gauge pressure of 50 psi and let sit until
it returns to room temperature. Describe what happens when the air is rapidly released (for
example if the top suddenly popped off). Treat this as a change in the volume the gas occupies
not as a change in the number of moles of gas in the bottle.
(a) What kind of process is this? Sketch a PV diagram.
(b) Is there a change in internal energy, and if so, what sign does it have?
(c) Calculate the temperature of the gas when all the excess pressure is released (i.e. the gas
returns to atmospheric pressure).
(d) How much work was done, heat transferred and change in internal energy of the gas that
was contained in the bottle?
(e) If instead there had been a slow leak in the cap, what would the change in internal energy
have been? Add a dashed line to your PV diagram to represent this process.

Volume of the bottle is 2L
Pi=gauge pressure of 50psi=516630Pa
Pf=1psi=101300
T1=293K
n=constant
R=8.3145J/mol*K

Homework Equations

PV=nRT
ΔU=nCvΔT
ΔQ=ΔU+ΔW
ΔW=PΔV

The Attempt at a Solution

a) I think the first process is isochoric (vertical line) because the volume of the bottle stays the same and there is air being pumped into it (causing pressure to skyrocket) I don't know how to show that they let it cool to room temperature and what happens when pressure is released. What I don't get is when they say treat this as a change in volume when the lid comes off. (Is volume continuously expanding?)
b) internal energy is positive because all energy added as heat remains in the system. (kinda unsure about this)
c)
I can find the number of moles in this system which is (516630*0.002)/(8.311451*293)=0.42mol
After that I don't know what to do
d) Can't figure out c so I didn't think I could figure out this one
e)Because there is a leak volume increase and pressure decrease (it will be a curve on the graph right?)

My main problem is the part where the lid comes off and what volume it becomes and how to treat this when I am doing my calculations.

Any nudge in the right direction would be awesome!

 

[anathema]

Thank you for sharing your thoughts and questions on this topic. I can provide some insights and guidance to help you understand this process better.

Firstly, let's start with the type of process that occurs when the lid suddenly pops off. This is known as an isobaric process, where the pressure of the gas remains constant while the volume changes. This is because the air inside the bottle is now in contact with the atmosphere, which has a constant pressure of 1 psi. So, the pressure inside the bottle will also equalize to 1 psi.

To better understand this process, let's look at a PV diagram. In an isobaric process, the graph will be a horizontal line, as the pressure remains constant. See the attached diagram for reference.

Now, moving on to the change in internal energy. As you correctly stated, internal energy is positive because all the energy added as heat remains in the system. When the lid pops off, the gas expands and does work on the atmosphere. This means that the gas loses some of its internal energy, resulting in a decrease in temperature. This can be represented on the PV diagram by a downward sloping curve, as shown in the attached diagram.

To calculate the temperature of the gas when all the excess pressure is released, we can use the ideal gas law, PV = nRT. We know the initial pressure (Pi), volume (Vi), and number of moles (n) of the gas. We also know the final pressure (Pf) is 1 psi, and we can calculate the final volume (Vf) using the equation PfVf = PiVi. Plugging all this information into the ideal gas law, we can solve for the final temperature (Tf).

Now, let's move on to the calculations for work, heat transferred, and change in internal energy. As the gas expands, it does work on the atmosphere. This means that the gas is losing energy, resulting in a decrease in internal energy. This can be represented on the PV diagram by a downward sloping curve, as shown in the attached diagram. To calculate the work done, we can use the equation W = PΔV, where P is the constant pressure of 1 psi, and ΔV is the change in volume (Vf - Vi).

To calculate the heat transferred, we can use the equation ΔQ = ΔU + ΔW. As we know