# Isothermal Expansion: Work and Heat

1. Dec 9, 2008

### kristibella

1. The problem statement, all variables and given/known data
I have most of these figured out (it is 4 and 5 that are causing me problems) but I figured that I may as well post the answers that I have already gotten because they may be relevant. The answers that I got are in bold.

1. Which of the curves in the figure represents an isothermal process?
A. B. C. D.

2. Calculate the change in internal energy DeltaU of the gas as it expands isothermally from V0 to the final volume fvV0. Here, fv is the ratio of final volume to initial volume; fv=4.0 in this problem.
DeltaU = 0

3. The pressure, temperature and volume of a system change differently during the process of isothermal expansion. Indicate whether each quantity increases (I), remains constant (C), or decreases (D).
Your answers should describe the changes of pressure, temperature, and volume (in that order). Separate your answers with commas (e.g., I,C,D means pressure increases, temperature remains constant, and volume decreases).
D,C,I

4. What is the work W done by the gas as it expands isothermally from V0 to fvV0?
Express the work done in terms of p0, V0, and fv. Use ln for the natural logarithm.

5. Calculate the quantity of heat Q that must be delivered to the gas as it expands from V0 to fvV0.
Express your answer in terms of p0, V0, and fv. Use ln for the natural logarithm.

2. Relevant equations
W = p(DeltaV) (At least, I hope so. It was briefly mentioned in lecture).
Q = W + (DeltaU)

3. The attempt at a solution
4. Work equals p(DeltaV), so for this problem, I said that W= p0(fvV0-V0). Mastering Physics said that "the answer would be correct if the pressure of the gas remained constant as the volume increased. Use the ideal gas law to determine how the pressure is changing as the volume changes." This confused me.

5. I think that this depends on the answer to number four. I think that it is (ANSWER TO FOUR) + 0.

2. Dec 9, 2008

### G01

For 4:

As Mastering Physics says, your answer would be correct if the pressure remained constant. That is the only time when $W=P\Delta V$ is true. Otherwise you must use the more general form:

$$W = \int P(V) dV$$

HINT: Use the ideal gas law to get P as a function of V.

For 5:

You do indeed need the answer of #4 to do number 5.

HINT: As you said, the internal energy of the gas doesn't change. So, where does the energy used to do the work come from?

3. Dec 9, 2008

### kristibella

I'm sorry, what does $$W = \int P(V) dV$$ mean? I know that is something with integral calculus but I haven't taken that class yet so I'm not exactly sure what it means.

4. Dec 9, 2008

### turin

You are in for some problems.

5. Dec 9, 2008

### Andrew Mason

You were probably given the relationship: $W = nRT\ln{(V/V_0)}$. That relationship is derived from the calculus integration that GO1 gave you.

For 5., you just have to know the relationship between W and Q if the temperature (hence internal energy) remains constant.

AM

6. Dec 9, 2008

Indeed. The only way that I know of to generate the proper formula for work is to integrate the function p(V)dV. However, if you do not know how to integrate, I am not sure how you are expected to know this unless you were asked to memorize the formulas for polytropic processes.

Casey

I see that Andrew gots you covered.

7. Dec 9, 2008

### kristibella

This class is supposed to be Trig based, Calc is not a prerequisite. This isn't the first time that I've been expected to know calculus, though.

8. Dec 9, 2008

### turin

Do you have a book? How does it tell you you to do this? I.e. how does the book tell you to calculate the work done by a gas?

9. Dec 9, 2008

### kristibella

According to the book, which was written by the professor, Q = W in an isothermal process. This was all that I could find on isothermal processes.

10. Dec 10, 2008

### turin

I am so sorry. I only had that horrible experience twice in my academic career. I can't stand professors that are so full of themselves that they think they need to write their own text for such a standard subject. Tell me you didn't have to go pick it up as a package of papers from Kinkos or something.

Well, that is true, but you usually use that to determine Q once you have calculated W. Unfortunately, that does not help you, because you still don't know how to calculate W. I guess the only thing you can do is what Andrew said. It may also be iluminating to realize that you can find the work done by the gas as the "area" under the curve on your plot between the two volumes. Of course, that won't give you an exact numerical answer, but it gives you a somewhat intuitive way to think about it. When the volume changes to the right on the graph (increases), the work is done BY the gas, and when the volume changes to the left, the work is done ON the gas. This graphical interpretation applies to any process that maintains a well-defined pressure (I think they call that "quasistatic"). In particular, you can use this basic idea to determine if the work done is greater than or less than for each of the four processes on your graph.

Last edited: Dec 10, 2008
11. Dec 10, 2008

### G01

Sorry. I got distracted by a research paper and haven't had time to check back on this thread.

This seems to be a moot point now, but I'll back up Andrew:

If you don't know calculus, the best you can do is use his formula and take it on faith that it is indeed true. That formula, as he said, is the result of computing the integral I posted.

You professor is correct that for an isothermal process, W=Q. So,use your answer from 4 to find the answer to five from this relation.

12. Dec 11, 2008