Thermodynamics Again HELP

1. Jun 21, 2006

cukitas2001

Hey guys, im understanding a bit more thermodynamics but ive been stumped once again on two problems:

1) A quantity of air is taken from state a to state b along a path that is a straight line in the pV-diagram

If the volume and pressure in state a are 7.40×10−2 m^3 and 1.08×10^5Pa , and those in state b are 0.115 m^3 and 1.32×10^5 Pa , what is the work W done by the gas in this process? Assume that the gas may be treated as ideal.

Is temperature constant here? I'm thinking so because of the line increasing at a constant rate. Do i need to apply $$W = \int p dV$$ with limits of $$V_2$$ and $$V_1$$ (don't know how to insert limits into the integral symbol on latex)

Any ideas?

2) A thermodynamic system is taken from state a to state c in the figure along either path abc or path adc. Along path abc the work done by the system is 450 J. Along path adc, is 120 J. The internal energies of each of the four states shown in the figure are Ua=150 J, Ub=240 J, Uc=680 J, Ud=330J.

A) Calculate the heat flow Q for the process ab.

I dont know how to go about this - work i can deal with on the pV diagrams but Q i don't know. Any help please?

2. Jun 21, 2006

Andrew Mason

Since $$W = \int_{V_1}^{V_2} p dV$$, = area under graph, does it matter how n changes in calculating W?
If P/V = constant and V = P/nRT, then what can you say about T and n?

AM

3. Jun 21, 2006

Andrew Mason

Use the first law of thermodynamics:

$$\Delta Q = \Delta U + \Delta W$$ where $\Delta W$ is the work done by the gas.

Is any work done by the gas in going from a to b?

AM

4. Jun 21, 2006

cukitas2001

I dont follow the first problem thinking

5. Jun 21, 2006

cukitas2001

no since its isochoric?

so then Q = 150-240 = -90 J....Heat is leaving the system?
or is it 240-150? for change being final minus initial so Ub-Ua?

Last edited: Jun 21, 2006
6. Jun 21, 2006

Andrew Mason

Is n a factor in the integral $\int PdV$ ? Do we have enough information to evaluate that integral (the area under the graph) ? Work out the expression for that area in terms of Pa, Pb, Vb and Va.

AM

7. Jun 21, 2006

cukitas2001

ok while im tryin to digest your advice for problem one....on problem two what would the work be from b to c ? would the intergral thing come up here again?

8. Jun 21, 2006

cukitas2001

substituting p=nRT/V into the integral wouldn't the constants nRT come out and only integration on p and V be done since p and v vary?

Could you perhaps elaborate some on you method to the first problem and start me off a bit?

Last edited: Jun 21, 2006
9. Jun 21, 2006

Tom Mattson

Staff Emeritus
He's trying to get you to see that you don't need to actually do an integral to get the area under the curve. You can break the region up into basic geometric shapes whose area formulae should be well known to you.

10. Jun 21, 2006

cukitas2001

ok so it should then be teh area of the triangle right? i tried doing:
W=(0.5)*(Vb-Va)*(Pb-Pa) and im getting 492J but its not right...what did i misinterpret?

11. Jun 21, 2006

Andrew Mason

You are omitting the rectangular portion below Pa.

AM

12. Jun 21, 2006

cukitas2001

ooooooooh ok got it now and ifgured out the work done by the horizontal line on the second problem...thanks for the help...

Hey Andrew, you think you could walk me through the mental process you were trying to show me i wanna try to understand it.

13. Jun 21, 2006

Andrew Mason

Which part?

I was trying to get you to realize that if you know P and V, you can determine the work done regardless of how n or T changes. Work is the area under the PV graph.

To your question on how temperature varies (which you do not have to know to find the work in this question), you can see that since P/V is constant and V = nRT/P, then nRT must be constant. So T must vary inversely as n (T = k/nR) where k = P/V

To your question on the heat flow, I was just trying to point out that if work is 0 (as it is from a to b, since there is no area under the PV graph from a to b) the heat flow is the difference in internal energy between b and a: Ub - Ua. In this case the change in internal energy from a to b is positive (increase) so heat flow is into the gas.

AM