PV work???

[MedGrl@2022]

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I just wanted to verify w=PdV not w=dPV. Therefore, volume must change in order for there to be work. There can be a change in pressure and a change in volume. But no work is done if there is a change in pressure but NO change in volume.

In order for their to be work there needs to be a change in volume. Because NO change in volume would mean that NO work has been done even if there is a change in pressure.

Please verify this.

Thank you,

Verónica

 

[cheesier]

You would never calculate work as dPV, however, there is such a thing as nonexpansion work. For instance, a combustion reaction carried out in a fixed volume container does work. Any time there is a change in the standard reaction gibbs energy, work is being done either by or on the system.

But if you're just talking about an ideal gas in a container, then the only kind of work is expansion work.

 

[milski]

I just wanted to verify w=PdV not w=dPV. Therefore, volume must change in order for there to be work. There can be a change in pressure and a change in volume. But no work is done if there is a change in pressure but NO change in volume.

In order for their to be work there needs to be a change in volume. Because NO change in volume would mean that NO work has been done even if there is a change in pressure.

Please verify this.

Thank you,

Verónica

Yes, that is correct. Work for constant pressure will be P * dV. For non-constant pressure the formula becomes more complicated but it still holds that dV=0 is the same as no work done.

You would never calculate work as dPV, however, there is such a thing as nonexpansion work. For instance, a combustion reaction carried out in a fixed volume container does work. Any time there is a change in the standard reaction gibbs energy, work is being done either by or on the system.

But if you're just talking about an ideal gas in a container, then the only kind of work is expansion work.

If it's just a combustion in a chamber all the added energy will be released as heat. You need some sort of movement or a chemical/electrical gradient to do non-expansion work. For MCAT purposes you can safely assume a simple mechanical work is considered and ΔV=0 means that W=0.

 

[MedPR]

Yes, that is correct. Work for constant pressure will be P * dV. For non-constant pressure the formula becomes more complicated but it still holds that dV=0 is the same as no work done.



If it's just a combustion in a chamber all the added energy will be released as heat. You need some sort of movement or a chemical/electrical gradient to do non-expansion work. For MCAT purposes you can safely assume a simple mechanical work is considered and ΔV=0 means that W=0.

Still killin it in the MCAT forum I see.

 

[cheesier]

If it's just a combustion in a chamber all the added energy will be released as heat. You need some sort of movement or a chemical/electrical gradient to do non-expansion work. For MCAT purposes you can safely assume a simple mechanical work is considered and ΔV=0 means that W=0.

Work is just energy put to use. However, in a combustion reaction like I described you generate heat but you also include the change in entropy in calculating work. I've got my pchem textbook in front of me and it says work(maximum, additional) = dG at constant temperature and pressure. I doubt you need to know that for the mcat though.

 

[milski]

Work is just energy put to use. However, in a combustion reaction like I described you generate heat but you also include the change in entropy in calculating work. I've got my pchem textbook in front of me and it says work(maximum, additional) = dG at constant temperature and pressure. I doubt you need to know that for the mcat though.

We mostly agree, except what you are saying is not applicable for the topic, aka PV work. The ΔG additional work in the equation above is for the case when T and P are constant. In the general case G=U-TS+PV. If you are allowed to change any of these, you can have a change in G without doing any work. For the combustion chamber case, you need some sort of "contraption" inside it which can put the increased temperature or pressure to use. If the chamber is just a fixed sized container containing the ignited substance, there will be no work done to/by the system.

 

[milski]

Still killin it in the MCAT forum I see.

One more year!

I hear you've changed a lot and mellowed out? 😛

 

[cheesier]

We mostly agree, except what you are saying is not applicable for the topic, aka PV work. The ΔG additional work in the equation above is for the case when T and P are constant. In the general case G=U-TS+PV. If you are allowed to change any of these, you can have a change in G without doing any work. For the combustion chamber case, you need some sort of "contraption" inside it which can put the increased temperature or pressure to use. If the chamber is just a fixed sized container containing the ignited substance, there will be no work done to/by the system.

It's applicable. The OP said:

"In order for their to be work there needs to be a change in volume. Because NO change in volume would mean that NO work has been done even if there is a change in pressure."'

This is false because even with constant P, V, and T, work can be done (but like I said, I seriously doubt the MCAT would ask you about it). You're right that nonexpansion work has to be harnessed somehow, but so does expansion work. That's why free expansion doesn't do work even though there's a change in volume. Pointing out that you need some contraption to actually use the energy doesn't address the fact that you can indeed do work without changing P, V, or T. Your G equation is also not right. You need to take the differential, and again, my pchem textbook agrees with what I stated prior. dG = w(additional, nonexpansion), meaning a change in G implies that you have energy that can be used for work.

 

[milski]

It's applicable. The OP said:

"In order for their to be work there needs to be a change in volume. Because NO change in volume would mean that NO work has been done even if there is a change in pressure."'

This is false because even with constant P, V, and T, work can be done (but like I said, I seriously doubt the MCAT would ask you about it). You're right that nonexpansion work has to be harnessed somehow, but so does expansion work. That's why free expansion doesn't do work even though there's a change in volume. Pointing out that you need some contraption to actually use the energy doesn't address the fact that you can indeed do work without changing P, V, or T. Your G equation is also not right. You need to take the differential, and again, my pchem textbook agrees with what I stated prior. dG = w(additional, nonexpansion), meaning a change in G implies that you have energy that can be used for work.

I disagree with a lot of this but there's not point in discussing it further.

 

[cheesier]

I disagree with a lot of this but there's not point in discussing it further.

LOL, you can't disagree with math but ok man. Honestly, if you didn't take pchem you wouldn't have learned it.

Check this out from your library if you're curious.

 

[advair250]

Why do you disagree with what cheesier said, don't you cover physical chemistry in your gen chem lol? But yeah, I suppose you wouldn't go into much depth in freshman sciences, i'd suggest physical chem as well if you haven't taken it