pressure, work

[chiddler]

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If you increase pressure of a gas at constant volume, how do you find the work done on that gas?

Say it's at 2 m^3 and it goes up 200 to 400 pascals.

 

[SaintJude]

Doesn't work = 0? Work is defined as W= PΔ V and since delta V is zero, W should also be zero.

 

[dmf2682]

Doesn't work = 0? Work is defined as W= PΔ V and since delta V is zero, W should also be zero.

Edit: Sorry misread your questions!

I dunno Jude. You're increasing the energy of the fluid so there's work being done on it. Unit wise, I would say the work is Delta p times v. 2*200 is 400 j.

 

[chiddler]

Stjude you're right. It's 0. I just found an EK question that asks this + answer.

Ok then my question is a conceptual why?

 

[milski]

Increasing only the pressure would increase the temperature. Keeping the temperature constant means that you are cooling down the system so whatever you are adding as energy by increasing the pressure you are losing as cool down.

 

[chiddler]

Increasing only the pressure would increase the temperature. Keeping the temperature constant means that you are cooling down the system so whatever you are adding as energy by increasing the pressure you are losing as cool down.

Can't the same thing be said if work is done to a system? The work that is done is energy that is removed from another system.

 

[milski]

Can't the same thing be said if work is done to a system? The work that is done is energy that is removed from another system.

That's not what I'm saying. I am talking about the same system - energy is added as the pressure is increased and lost as the temperature is lowered back to what it was.

 

[chiddler]

That's not what I'm saying. I am talking about the same system - energy is added as the pressure is increased and lost as the temperature is lowered back to what it was.

i'm sorry, I don't understand. could you please rephrase it for me?

 

[milski]

i'm sorry, I don't understand. could you please rephrase it for me?

Of course you don't, for some reason I was thinking about the case where the volume is being decreased to raise the pressure. Actually, 'thinking' is probably not the right word for what I was doing. 😡

There are two ways to increase the pressure for a constant volume:

- increase the amount of matter - that's not work, that's just making the system larger.

- raise the temperature - again, adding heat is not work, it's just adding heat

 

[chiddler]

Of course you don't, for some reason I was thinking about the case where the volume is being decreased to raise the pressure. Actually, 'thinking' is probably not the right word for what I was doing. 😡

There are two ways to increase the pressure for a constant volume:

- increase the amount of matter - that's not work, that's just making the system larger.

- raise the temperature - again, adding heat is not work, it's just adding heat

lol that's ok, i appreciate your responses regardless.

Heat is energy. When work is done, energy is transferred. So if heat is added to a system, i'd expect work to be done in order for the energy to be added. Why is this not the case?

 

[milski]

lol that's ok, i appreciate your responses regardless.

Heat is energy. When work is done, energy is transferred. So if heat is added to a system, i'd expect work to be done in order for the energy to be added. Why is this not the case?

You can transfer energy by doing work or transferring heat. The two are different things. Work involves some sort of displacement, even if it's only statistical, like the case of ΔV. Changes in heat involve changes in the internal state of the matter - kinetic/vibrational energy of its particles.

 

[chiddler]

You can transfer energy by doing work or transferring heat. The two are different things. Work involves some sort of displacement, even if it's only statistical, like the case of ΔV. Changes in heat involve changes in the internal state of the matter - kinetic/vibrational energy of its particles.

Oh. Thus the U = Q + W equation having independent variables for both heat and work.

thanks that helped a lot.

 

[milski]

Oh. Thus the U = Q + W equation having independent variables for both heat and work.

thanks that helped a lot.

Yes. Heat is a slacker and lacks organization to be considered real work. 😉

 

[dmf2682]

Yes. Heat is a slacker and lacks organization to be considered real work. 😉

Cute.