Gas at a constant pressure: using Cv.dT correctly

[chopnhack]

Hello, my questions is not so much homework, but a request for a definition. When we use Cv*dT to solve for dU (internal energy) in a constant pressure example, what is the order of the temperatures entered into dT?
Is one to assume its final temp minus initial temp? I ask because it leads to sign convention and I want to make sure I understand it properly.

Thanks!

 

[TSny]

Yes, dT represents final temperature minus initial temperature for an infinitesimal change in temperature.

Positive dT corresponds to a temperature increase, which also corresponds to an increase in internal energy of the gas. It doesn't matter whether the pressure is constant or not.

 

[chopnhack]

Thank you, that helped me a lot!

Could you point me in the right direction for the last part of this problem? I tried to simply sum up the dQ's, dU's and dW's, but noticed that there was overlap where in some cases dU = dW... I then thought out the process and how the gas starts at a certain temperature and pressure, the gas loses heat while expanding and regains said heat while contracting. I thought that they might cancel each other out. The gas returns to its original starting point, temperature and pressure - no net W done.

Thanks in advance.

 

[Chestermiller]

Thank you, that helped me a lot!

Could you point me in the right direction for the last part of this problem? I tried to simply sum up the dQ's, dU's and dW's, but noticed that there was overlap where in some cases dU = dW... I then thought out the process and how the gas starts at a certain temperature and pressure, the gas loses heat while expanding and regains said heat while contracting. I thought that they might cancel each other out. The gas returns to its original starting point, temperature and pressure - no net W done.

Thanks in advance.

In step 1, $\Delta U$ is negative, since the temperature decreased. If you correct this, then you will find that $\Delta U$ for the cycle is equal to zero. You should also then find that W for the cycle is equal to Q for the cycle.

 

[chopnhack]

In step 1, $\Delta U$ is negative, since the temperature decreased. If you correct this, then you will find that $\Delta U$ for the cycle is equal to zero. You should also then find that W for the cycle is equal to Q for the cycle.

I was close!
I missed the fact that the calculation was final temperature minus initial temperature for dT which yielded the negative value for dU and dW for process one.

Process 1: Adiabatic

dQ = 0 No heat added in an Adiabatic expansion, the gas expands using internal energy.

dU = dW = Cv×dT = 12.471 JK-1 × (389K-588K) --> -2,482 J

But now that I look at the other parts, my process for isobaric would need to yield a positive value to have the whole system balance out with 993J Q and W
My understanding of the equation was that final volume minus initial volume in this case will yield a negative value...
Where have I gone wrong?

Process 2: Isobaric

W = P(V2-V1)--> 3.59x10^4Pa (0.0484m3- 0.0899m3) --> -1,490 J work done on the gas - final volume is less.

dU = Cv×dT à 12.471 JK-1 × (209.4K-389K) = -2,239 J

dQ = -2,239 J + -1,490 J = -3,729 J

 

[Chestermiller]

I was close!
I missed the fact that the calculation was final temperature minus initial temperature for dT which yielded the negative value for dU and dW for process one.

Process 1: Adiabatic

dQ = 0 No heat added in an Adiabatic expansion, the gas expands using internal energy.

dU = dW = Cv×dT = 12.471 JK-1 × (389K-588K) --> -2,482 J

No. $$\Delta U=Q-W$$so, if Q = 0, $$W=-\Delta U=+2482 J$$

But now that I look at the other parts, my process for isobaric would need to yield a positive value to have the whole system balance out with 993J Q and W
My understanding of the equation was that final volume minus initial volume in this case will yield a negative value...
Where have I gone wrong?

Process 2: Isobaric

W = P(V2-V1)--> 3.59x10^4Pa (0.0484m3- 0.0899m3) --> -1,490 J work done on the gas - final volume is less.

dU = Cv×dT à 12.471 JK-1 × (209.4K-389K) = -2,239 J

dQ = -2,239 J + -1,490 J = -3,729 J

Process 2 calc is correct.

 

[chopnhack]

No. $$\Delta U=Q-W$$so, if Q = 0, $$W=-\Delta U=+2482 J$$

Process 2 calc is correct.

Ok, I see the manipulation, simply restating the Q=U+W equation, solve for U instead. What would lead me to do that though? The lecture notes showed us many derivations and the one that came to mind when solving this problem was dU = dW = Cv×dT (for adiabatic process). Is this equation wrong? Forgive my questioning, but I am trying to understand the why. I don't want to simply plod through the mechanics of the math.

(Looking at it from a strict definition standpoint it makes sense - for an adiabatic process - internal energy changes as much as work done, but since the work done come from the gas itself, the value changes!)

 

[Chestermiller]

Ok, I see the manipulation, simply restating the Q=U+W equation, solve for U instead. What would lead me to do that though? The lecture notes showed us many derivations and the one that came to mind when solving this problem was dU = dW = Cv×dT (for adiabatic process). Is this equation wrong? Forgive my questioning, but I am trying to understand the why. I don't want to simply plod through the mechanics of the math.

There are two sign conventions used in the literature for the first law: $$\Delta U=Q-W$$where W is the work done by the system on the surroundings, and $$\Delta U=Q+W$$whereW is the work done by the surroundings on the system? Which sign convention are you using?

 

[chopnhack]

There are two sign conventions used in the literature for the first law: $$\Delta U=Q-W$$where W is the work done by the system on the surroundings, and $$\Delta U=Q+W$$whereW is the work done by the surroundings on the system? Which sign convention are you using?

Can we use dW = -dU?

Process 1: Adiabatic

dQ = 0 No heat added in an Adiabatic expansion, the gas expands using internal energy.

dW = -Cv×dT = 12.471 JK-1 × (389K-588K) à +2,482 J

dU = -dW = -2,482 J

 

[Chestermiller]

Can we use dW = -dU

Only if Q = 0. Otherwise $$\Delta U=Q-W$$and$$W=Q-\Delta U$$

(Please don't use d's when you mean $\Delta$'s or, in the case of path-dependent quantities like heat and work, capitalized symbols Q and W).

 

[chopnhack]

Only if Q = 0. Otherwise $$\Delta U=Q-W$$and$$W=Q-\Delta U$$

(Please don't use d's when you mean $\Delta$'s or, in the case of path-dependent quantities like heat and work, capitalized symbols Q and W).

Got it! only for adiabatic. Sorry, I was using what the professor taught us. He did mention not to call it 'd' but delta. Maybe he had a hard time with his software ;-)
Thank you for helping me understand this better. I was having fits with it!