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First Law of Thermodynamics - Correct Formulation?

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Homework Statement



A gas expands from 1L at 1atm to 3L. Assume P is directly proportional to V.
This is a simple question from my second lecture on Thermal Physics, and I have found a way to solve it after my professor made it very confusing in class. My question is more about the formulation of the first law of thermodynamics than this problem in particular. I am confused as to what formulation is valid under what assumptions, rather what assumptions make Eq. 1 (below) valid?




Homework Equations



My professor in class stated the first law of thermodynamics as such:

[tex]\Delta E = q + w[/tex] Eq. (1)

Whereas in Baierlein, our course book, it is stated as:

[tex]q = \Delta E + w [/tex] Eq. (2)


The Attempt at a Solution



I have a reference book (Theoretical Physics by George Joos, Dover Edition) that addresses this discrepancy. They present Eq. (1) as one form of the First Law of Thermodynamics under the conditions that all forms of energy are considered to be positive. My thought in this case is that the change in internal energy would not be negative even when the gas expands, and it is up to whomever is using the equation to determine the sign using physical intuition.
Secondly, when they discuss "a change in volume accompanying an external pressure p" they derive Eq. (2) as a special case of the first law, wherein the quantities of the different forms of energy can be either positive or negative. This seems much more intuitive to me.
I believe I was able to solve the problem using the second equation and taking the work done by a gas on its surrounding to be positive (and compression to be negative work), but I cannot see how to solve the problem--or even how exactly to define the problem--if the first formulation of the law is also correct. I have attached a pdf of my solution and the reference book I used. It would be helpful if someone could solve the problem using Eq. (1) and the proper assumptions.
 

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  • #2
Andrew Mason
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Homework Statement



A gas expands from 1L at 1atm to 3L. Assume P is directly proportional to V.
This is a simple question from my second lecture on Thermal Physics, and I have found a way to solve it after my professor made it very confusing in class. My question is more about the formulation of the first law of thermodynamics than this problem in particular. I am confused as to what formulation is valid under what assumptions, rather what assumptions make Eq. 1 (below) valid?




Homework Equations



My professor in class stated the first law of thermodynamics as such:

[tex]\Delta E = q + w[/tex] Eq. (1)

Whereas in Baierlein, our course book, it is stated as:

[tex]q = \Delta E + w [/tex] Eq. (2)
In 1., W is the work done ON the gas. In 2., W is the work done BY the gas. Text books used to use 1. but now most use 2. Just a different way of stating the law.

In 1. the change in internal energy of the system is the sum of the energies (heat flow and work) transferred to the system. That makes sense. But the problem was with the sign of W that always led to confusion. Work done ON the gas is [itex]-P\Delta V[/itex]

I believe I was able to solve the problem using the second equation and taking the work done by a gas on its surrounding to be positive (and compression to be negative work), but I cannot see how to solve the problem--or even how exactly to define the problem--if the first formulation of the law is also correct. I have attached a pdf of my solution and the reference book I used. It would be helpful if someone could solve the problem using Eq. (1) and the proper assumptions.
In your problem, why do you think that P changes? Does this expansion not occur at constant P = 1 atm?

AM
 
  • #3
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Okay, so that's the only difference thanks a bunch. I think my professor had mentioned that she was using a difference book for her notes than we had read. She hadn't been able to reconcile the difference by the end of class though--it just became a big mess.

Regarding why P changes, I think the confusion comes only from a poor statement of the problem. I know that in general for an ideal gas there is an inverse relationship between P and V, but the example was supposed to be more primitive than that and was just supposed to illustrate that the work was the area under the curve [tex] P(V) [/tex]. We are supposed to assume that the gas expands in such a way that the pressure remains directly proportional to the volume. Is this not possible, or can it be achieved in practice by varying other parameters?
 
  • #4
Andrew Mason
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Regarding why P changes, I think the confusion comes only from a poor statement of the problem. I know that in general for an ideal gas there is an inverse relationship between P and V, but the example was supposed to be more primitive than that and was just supposed to illustrate that the work was the area under the curve [tex] P(V) [/tex]. We are supposed to assume that the gas expands in such a way that the pressure remains directly proportional to the volume. Is this not possible, or can it be achieved in practice by varying other parameters?
Ok. I see that is given in the problem that [itex]P\propto V[/itex].

PV=nRT is still the relationship between P, V, and T in an ideal gas.

If the pressure is proportional to V in this process (ie. heat flows in to the gas as it expands), then P/V = constant or P = kV. Use the relationship between P, V and T to determine k.

To calculate the work, you must use:

[tex]W = \int_{V_i}^{V_f} PdV = \int_{V_i}^{V_f} kVdV[/tex]

AM
 
  • #5
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So, if we use the ideal gas law, and divide by [tex]V^2[/tex]...

[tex] \frac{P}{V} = \frac{Nk_BT}{V^2} = k [/tex]

So we see that, if [tex]P \propto V [/tex], the expansion cannot be isothermal; moreover there must be a specific relationship between T and V? I don't really think that makes much sense that the proportionality constant between P and V includes V itself--it seems like that makes [tex]P \propto V^{-1}[/tex]--but I don't see a way around that. How can an ideal gas have [tex]P \propto V[/tex] then; am I just wrong in doubting that V can be included in the constant [tex]k[/tex]?
 
  • #6
Andrew Mason
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So, if we use the ideal gas law, and divide by [tex]V^2[/tex]...

[tex] \frac{P}{V} = \frac{Nk_BT}{V^2} = k [/tex]

So we see that, if [tex]P \propto V [/tex], the expansion cannot be isothermal; moreover there must be a specific relationship between T and V? I don't really think that makes much sense that the proportionality constant between P and V includes V itself--it seems like that makes [tex]P \propto V^{-1}[/tex]--but I don't see a way around that. How can an ideal gas have [tex]P \propto V[/tex] then; am I just wrong in doubting that V can be included in the constant [tex]k[/tex]?
If temperature does not change [itex]P \propto 1/V[/itex]

But if T increases as the square of volume: [itex]T \propto V^2[/itex] then, since P is proportional to T but inversely proportional to V, (P=nRT/V), P will be proportional to V.

Now since P/V = k it is just a matter of finding the value of that constant. All you need is one value of P, V and T that fit the relationship (since the constant is the same for all values). We know that the initial values fit that relationship, so:

[tex]P_0 = kV_0 = nRT_0/V_0[/tex]

so:

[tex]k = nRT_0/V_0^2 = P/V[/tex]

For the relationship [itex]k = P/V[/itex] to hold,

[tex]T = PV/nR = kV^2/nR = (nRT_0/V_0^2)V^2/nR = V^2(T_0/V_0^2)[/tex]

AM
 
  • #7
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Okay, gotcha! Thanks a bunch
 

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