Calculating Work Done on Surroundings for Thermodynamic Process

[Philip Wong]

Homework Statement

Two moles of an ideal monoatomic gas trebles its initial volume is an isobraic expansion from state A to state B. The gas is then cooled ischorically to state C and finally compressed isothermally until it returns to state A.
The molar gas constat is: R = 8.314 J mol^-1 K^-1.
Boltzmann's constant: 1.38*10^-23JK^-1
state B temperature = 803K
state A temperature = 268K

What is the work done on the surroundings in each of the following processes:
(i) A > B
(ii) B > C
(iii) C > A

Homework Equations

(i)W= (intergration)n.R.T/V.dv
= n.R.T(B).(intergration) 1/V. dv
= n.R.T(A).ln(V(B)/V(A))

(ii) W = 0

(iii) W=n.R.(T(B)-T(A))

The Attempt at a Solution

(i) W = n.R.T(A).ln(V(B)/V(A))
=2*8.341*268*ln(2) = 3099J (which was wrong... correct answer should be 8900)

(ii) W= 0 as there is no work needed for cooling of the system

(iii) W=n.R.(T(A)-T(B))
2*8.341*(268-803) = -8924 (which was wrong... correct answer should be -4890)


can anyone tell me where did i do wrong?

thanks!

 

[vela]

Homework Statement

Two moles of an ideal monoatomic gas trebles its initial volume is an isobraic expansion from state A to state B. The gas is then cooled ischorically to state C and finally compressed isothermally until it returns to state A.
The molar gas constat is: R = 8.314 J mol^-1 K^-1.
Boltzmann's constant: 1.38*10^-23JK^-1
state B temperature = 803K
state A temperature = 268K

What is the work done on the surroundings in each of the following processes:
(i) A > B
(ii) B > C
(iii) C > A

Homework Equations

(i)W= (intergration)n.R.T/V.dv
= n.R.T(B).(intergration) 1/V. dv
= n.R.T(A).ln(V(B)/V(A))

(ii) W = 0

(iii) W=n.R.(T(B)-T(A))

The Attempt at a Solution

(i) W = n.R.T(A).ln(V(B)/V(A))
=2*8.341*268*ln(2) = 3099J (which was wrong... correct answer should be 8900)

This equation doesn't apply to an isobaric expansion. You can't just use random equations because you think you have values for all the variables; you need to make sure the assumptions that go into the derivations are met. The equation you used only applies to constant-temperature processes. (That's why you could pull T out of the integral in the derivation.) The fact that you had to arbitrarily choose one of the temperatures to plug into the equation should have been a clue that you were headed in the wrong direction.

You should have expressions for the work done for various processes. You need to find or, better yet, derive one that applies to the process that takes you from state A to state B. The key word for this part is isobaric. What does isobaric mean?

(ii) W= 0 as there is no work needed for cooling of the system

Right answer, wrong reason. A gas in an insulated container, for example, can cool only by expanding.

(iii) W=n.R.(T(A)-T(B))
2*8.341*(268-803) = -8924 (which was wrong... correct answer should be -4890)

can anyone tell me where did i do wrong?

thanks!

Same issue as with part (i), but this time, the process is isothermal.

 

[Philip Wong]

ha! strange really, because when question is formatted at: isothermal expansion from state A to state B, then compressed isobarically to state C, and finally heated ischorically until it returns to state A. these are the exact steps I took, and I got the answer correct (attempted a few question, so chance are I got it right by chance is extremely unlikely). but when it comes to cooled and before compression I use the same steps again but now is wrong.

anyways isobaric, as I recall it the formula to calculate work done is:
W = P*(delta) V

 

[Philip Wong]

for II)
W = 0, because cooling can only be done by expansion. since there is already expansion done, therefore no extra work is need for the gas to cool down?

 

[vela]

for II)
W = 0, because cooling can only be done by expansion.

No. For an ideal gas, a drop in temperature means the internal energy of the gas decreases. According to the first law, that energy has to go somewhere. It's either transferred out as heat or as work done by the gas on the surroundings.

since there is already expansion done, therefore no extra work is need for the gas to cool down?

You seem to be under the impression that a change is made, e.g. expansion from A to B, and then the effect (cooling from B to C) follows some time later. That's not how it works. Processes are assumed to proceed quasi-statically, that is, slowly enough so that the system has time to adjust and essentially remain in equilibrium. Any cooling due to expansion from A to B is already reflected in the temperature at B.

What does isochoric mean?

 

[Philip Wong]

isochoric meant that it is a constant volume process?
therefore things always stays at a equilibrium status no matter what change it has undergo? i.e. heat up, compress, or expaned?

 

[Philip Wong]

No. For an ideal gas, a drop in temperature means the internal energy of the gas decreases. According to the first law, that energy has to go somewhere. It's either transferred out as heat or as work done by the gas on the surroundings.

You seem to be under the impression that a change is made, e.g. expansion from A to B, and then the effect (cooling from B to C) follows some time later. That's not how it works. Processes are assumed to proceed quasi-statically, that is, slowly enough so that the system has time to adjust and essentially remain in equilibrium. Any cooling due to expansion from A to B is already reflected in the temperature at B.

What does isochoric mean?

So things happening are instantaneous (i.e. all happening at the same time?), but when I look at a PV graph, say for a otto engine I always get the impression that things are happening in chain progress (i.e. one process is completed before another happens) or did I understood it wrong?

 

[vela]

I have to admit I'm having a great deal of difficulty understanding what you're thinking. I'm not sure if you're just not expressing yourself clearly, but to me, some of your questions and statements seem to reveal really fundamental misconceptions about thermodynamics.

Let's start over. Here's what you said was the original problem:

Homework Statement

Two moles of an ideal monoatomic gas trebles its initial volume is an isobraic expansion from state A to state B. The gas is then cooled ischorically to state C and finally compressed isothermally until it returns to state A.
The molar gas constat is: R = 8.314 J mol^-1 K^-1.
Boltzmann's constant: 1.38*10^-23JK^-1
state B temperature = 803K
state A temperature = 268K

What is the work done on the surroundings in each of the following processes:
(i) A > B
(ii) B > C
(iii) C > A

Then later you say this:

ha! strange really, because when question is formatted at: isothermal expansion from state A to state B, then compressed isobarically to state C, and finally heated ischorically until it returns to state A. these are the exact steps I took, and I got the answer correct (attempted a few question, so chance are I got it right by chance is extremely unlikely). but when it comes to cooled and before compression I use the same steps again but now is wrong.

In the first problem, you have an isobaric expansion from A to B. In the second problem, you have an isothermal expansion from A to B. Why would you expect applying the exact same steps of the second problem would give you the correct answer for the first problem? They are different problems.

In the second problem, you got the correct answer because you applied a formula for an isothermal process and the process connecting A to B was an isothermal expansion. In the first problem, however, you tried to use the same formula even though the expansion from A to B was isobaric. The formula won't work here because the expansion is not isothermal. This is what I was getting at in my reply, yet you seemed to completely miss that.

As you noted later, for an isobaric process, the work is given by $W=P \Delta V$. Unfortunately, you're not given the pressure or volumes, so you might want to look for a more appropriate formula for the work done by an isobaric expansion (or derive it using the ideal gas law).

Once you get this part, we can move on to part (ii).

 

[Philip Wong]

I have to admit I'm having a great deal of difficulty understanding what you're thinking. I'm not sure if you're just not expressing yourself clearly, but to me, some of your questions and statements seem to reveal really fundamental misconceptions about thermodynamics.

Let's start over. Here's what you said was the original problem:

Then later you say this:

In the first problem, you have an isobaric expansion from A to B. In the second problem, you have an isothermal expansion from A to B. Why would you expect applying the exact same steps of the second problem would give you the correct answer for the first problem? They are different problems.

In the second problem, you got the correct answer because you applied a formula for an isothermal process and the process connecting A to B was an isothermal expansion. In the first problem, however, you tried to use the same formula even though the expansion from A to B was isobaric. The formula won't work here because the expansion is not isothermal. This is what I was getting at in my reply, yet you seemed to completely miss that.

As you noted later, for an isobaric process, the work is given by $W=P \Delta V$. Unfortunately, you're not given the pressure or volumes, so you might want to look for a more appropriate formula for the work done by an isobaric expansion (or derive it using the ideal gas law).

Once you get this part, we can move on to part (ii).

I'm terribly sorry, I think I've delieved my question wrong which confused you. What I meant was apart from the question I posted at the beinging of the post I was also given another set of question to work with, which is:

one mole of an ideal monoatomic gas doubles its initial volume in an isothermal expansio from state A to state B. The gas is then compressed isobarically to state C and finally heated isochorically until it returns to state A.

a) If state C corresponds to a pressure p=4 atm, and temperature T = 1069K, determine the temperature of the gas in state B.

I worked out temerature of B by:
P. V = nRT
SO (P.V)/T = n . R Where R is a constant
SO (P(B)*V(B))/T(B) = (P(C)*V(C))/T(C)
SO T(C) * (P(B)*V(B))/T(B) = P(C)*V(C)
SO T(C) * V(B))/T(B) = (P(C)*V(C)) * T(B)
SO T(B) = T(C) * P*(B)/P(C) * V(B)/V(C)
Therefore TB = 1069 X 1 X 2 (2 because volume doubles)
= 2138 K

b)
i) A > B
For this part I use the following equation:
I know W = delta U, I also know U = P*V
Therefore W = P/V, BUT I know P is calculated from n.R.T
SO W= (intergration)n.R.T/V.dv
SO n.R.T(B).(intergration) 1/V. dv
SO n.R.T(B).ln(V(B)/V(A))
n = 1 (one mole of ideal monoatomic gas)
R = 8.314
T(B) = 2138
ln(V(B)/V(A)) = ln(2)
Therefore n.R.T(B).ln(V(B)/V(A)) = 1*8.314*2138*ln(2) = 12321J

ii) B > C
W=n.R.(T(B)-T(C))
I use (T(B)-T(C)) because it is compress to state C from state B, when there is compression made temperature decrease (whearse if it is expand to state C from state B, I would use (T(B)+T(C))
=1*8.314*(2138-1069)
=8887.7J

iii) C > A
0

These are the steps I took for this particular question, which somehow I got it all right. Therefore I used these steps again for the question I posted originally in this forum. WHICH gives me a wrong answer... I don't know why...

 

[vela]

I know, and I've explained why you're getting the wrong answer. Did you even read my last post?

 

[Philip Wong]

I know, and I've explained why you're getting the wrong answer. Did you even read my last post?

is it the one about where you said:
In the first problem, you have an isobaric expansion from A to B. In the second problem, you have an isothermal expansion from A to B. Why would you expect applying the exact same steps of the second problem would give you the correct answer for the first problem? They are different problems.

In the second problem, you got the correct answer because you applied a formula for an isothermal process and the process connecting A to B was an isothermal expansion. In the first problem, however, you tried to use the same formula even though the expansion from A to B was isobaric. The formula won't work here because the expansion is not isothermal. This is what I was getting at in my reply, yet you seemed to completely miss that.

----------
So I should look for something that I could used to solve for isobaric, instead of isothermal. This is why when I took the same steps again from the second problem, and applied it to the first problem I got it wrong.

 

[vela]

So I should look for something that I could used to solve for isobaric, instead of isothermal. This is why when I took the same steps again from the second problem, and applied it to the first problem I got it wrong.

Again, in the second problem, A->B is isothermal; in the first problem, A->B is isobaric. Yet you're trying to use the same formula in both cases. The formula you used only works for an isothermal process, which is why you got the correct answer for the second problem; it doesn't apply to the case of an isobaric process, which is why you got the wrong answer in the first problem.

 

[Philip Wong]

now I'm lost, so how should I solve isobaric process?? I've been thinking about it for the pass few days and I don't seems to get my head around it

 

[vela]

Well, you said for an isobaric process, the work is given by W=PΔV. You need to figure out how to calculate that quantity.

You might want to seek help from your instructor, a teaching assistant, or a tutor. You seem to have some fundamental misconceptions that would be better cleared up by talking to someone one-on-one rather than posting on a forum like this.

 

[Philip Wong]

alright! sounds like a good idea! thanks for trying to help!