Adiabatic Work

[SSerenity]

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Im not grasping this concept entirely.

The question is, When the stop cock is opened, how much work is done by the GAS as the system reaches equilibrium?

The first law of thermo says dU = dQ + dW

Since this is an isolated system, there is no heat exchange with the environment. So this sets our dQ term = 0. Makes sense so far.

And the Total Internal Energy is directly proportional to temperature. Since the temperature stays the same, Our U term also is zero. Therefore the Work also is zero.

If the gas temperature stays the same...as soon as the stop cock is opened, the gas volume will increase. To counter act this, the pressure must then decrease, correct?

What of the Pressure Volume work done by the gas? If I was to plot Volume vs Pressure, I would have a line with a negative slop. The area under this line gives me a non-zero value.

Where does the PV work come into play?


Thanks!

 

[monkeyvokes]

Im not grasping this concept entirely.

The question is, When the stop cock is opened, how much work is done by the GAS as the system reaches equilibrium?

The first law of thermo says dU = dQ + dW

Since this is an isolated system, there is no heat exchange with the environment. So this sets our dQ term = 0. Makes sense so far.

And the Total Internal Energy is directly proportional to temperature. Since the temperature stays the same, Our U term also is zero. Therefore the Work also is zero.

If the gas temperature stays the same...as soon as the stop cock is opened, the gas volume will increase. To counter act this, the pressure must then decrease, correct?

What of the Pressure Volume work done by the gas? If I was to plot Volume vs Pressure, I would have a line with a negative slop. The area under this line gives me a non-zero value.

Where does the PV work come into play?


Thanks!

Why does the temperature stay the same? Doesn't it decrease ?
Adiabatic decrease in pressure and increase in volume should correspond to a temperature drop, I think. This is what occurs in the carnot cycle, but I'm not 100% sure of this stuff either. Hopefully others respond

 

[SSerenity]

I should have mentioned, this is an ideal gas. So there are no van der waals forces. If it was a real gas, then as the molecules get further apart their kinetic energy is converted into potential energy (hence the drop in temperature).

Apparently, even if this was a real gas, and there was a drop in temperature inside the gas it would still be adiabatic because our dQ does not change. Q = PE + KE (given our insulation was perfect)

 

[monkeyvokes]

I should have mentioned, this is an ideal gas. So there are no van der waals forces. If it was a real gas, then as the molecules get further apart their kinetic energy is converted into potential energy (hence the drop in temperature).

Apparently, even if this was a real gas, and there was a drop in temperature inside the gas it would still be adiabatic because our dQ does not change. Q = PE + KE (given our insulation was perfect)

Alright, how does this information relate to your initial question of where PV work comes in?

 

[SSerenity]

I don't think it does 🙁

 

[monkeyvokes]

I don't think it does 🙁

is the magnitude of work done related to the change in temperature? And the negative slope because the temperature decreases?

I think the sign can also change depending on how you frame it (work done by gas or work done on gas). Any other ideas..?

 

[sps27]

I was a bit confused about this myself, so had to do some search on the Internet. The figure above suggests Adiabatic Free Expansion of a gas. If you search for that in google (if you haven't done that already), you will find many links. From what I could understand after reading a little bit on this is, all 3 variables of the first law = 0 in the case of free expansion of a gas, be it ideal or real. For a real gas there will be a little dip in temp due to decreased pressure but still there is no work done because the expansion is free and nothing resisting it.

http://www.pha.jhu.edu/~broholm/l35/node6.html

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node33.html

 

[mehc012]

I always think of it in terms of the whole container, not the gas.

The average pressure in the container is constant. The volume of the container is constant. The number of molecules of gas is constant. The only thing that changes is the arrangement.

Ignoring the extreeeeeemely low probability of it actually happening, the first arrangement is one of the possible snapshots of the container of gas described the final PV=nRT variables.

It's an increase in local entropy, and doesn't require input work/energy.

 

[randymoss]

There should be no work done and since you said it's an isolated system, no change in heat, so dQ shouldn't change.

I think you're not drawing the negative slope of your graph far enough down below the x-axis. With pressure on the y-axis and volume on the x-axis, you're probably drawing the line down to where pressure = 0. Since the volume of the system doubles when the system is opened to the other chamber, you should continue drawing the negative sloping line below the x-axis the same distance it started above the x-axis (since this now represents a complete evacuation of the original chamber). This should give you a net of zero under the PV line.

 

[SSerenity]

There should be no work done and since you said it's an isolated system, no change in heat, so dQ shouldn't change.

I think you're not drawing the negative slope of your graph far enough down below the x-axis. With pressure on the y-axis and volume on the x-axis, you're probably drawing the line down to where pressure = 0. Since the volume of the system doubles when the system is opened to the other chamber, you should continue drawing the negative sloping line below the x-axis the same distance it started above the x-axis (since this now represents a complete evacuation of the original chamber). This should give you a net of zero under the PV line.

You are right, this is where I messed up. Thanks! 👍