# Ideal Gas Law + 1st Law of Thermodynamics

1. Dec 28, 2006

### twotaileddemon

Hi everyone ^^! I hope you had a great holiday season. I have a homework problem that I'm a little confused on. I will provide my answers/work as usual, and would like it if someone could check my work, or tell me if an explanation is wrong/needs more information. My teacher would like us to be very specific on these so I'm not sure if my explainations are well-explained enough. Though, I would like to say that I've been doing very well on my tests lately because I apply what I learn from everyone here on my tests and am finally starting to do really well ^^. Thanks.

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Solve the following problem. (Hint: Use the Ideal Gas Laws and the first law of thermodynamics) Be specific and support your responses.

Diagram: http://img.photobucket.com/albums/v696/talimtails/PP18.jpg

Four samples of ideal gas are each initially at pressure Po, a volume Vl, and a temperature To, as show on the diagram above. The samples are taken in separate experiments from this initial state to the final states I, II, III, and IV along the processes shown on the diagram.

a. One of the processes is isothermal. Identify this one and explain.
Process II is isothermal. The temperature remains constant and pressure varies directly with volume.

b. One of the processes is adiabatic. Identify this one and explain.
Process III is adiabatic because no heat is transferred during the process. Furthermore, adiabatic processes have a steeper inclination than isotherms because they lose more pressure during expansion.

c. In which process or processes does gas do work? Explain.
W = P∆V. In process IV volume remains constant and therefore there is no work. Thus, in processes I, II, and III work is done because there is a change in volume.

d. In which process or processes is heat removed from the gas? Explain.
∆U = Q – W.
In processes II, III, and IV heat is removed from the gas because there is a decrease from the initial temperature as shown by the decreasing slopes of the lines.

e. In which process or processes does the root-mean-square speed of the gas molecules increase? Explain.
Vrms = sqrt (3RT/Mm)
If the temperature decreases, the speed of the gas particles decreases as well. Therefore, because none of the processes experience an increase in temperature, there are no processes in which the root-mean-square speed of the gas molecules increase.

2. Dec 30, 2006

### twotaileddemon

If the diagram is too small I'll draw a bigger one

3. Dec 31, 2006

### twotaileddemon

Oh, and process II should end slightly lower on the Po/2 line exactly. :)

4. Dec 31, 2006

### Andrew Mason

With your correction of the graph (ending on the line P = P0/2), Process II must be the isothermal. If T is constant then $P = k/V$ where k = nRT0. The beginning and end points of II, only, fit that relationship.
Correct, although I am not sure how you are determining from the graph that no heat is transferred. Since $P = K/V^\gamma$ where $\gamma > 1$ it is similar P = k/V but P decreases more rapidly with volume.
Correct. The area under the graph is the work done. Only in process IV is this area 0.
Not quite. For the adiabatic (III), no heat is removed so this is the line of Q = 0. For lines above the adiabatic internal energy is higher so heat must be added (Q>0). Below the adiabatic, Q<0 (heat removed). Which lines are above the adiabatic?

Not quite. What about Process I? How does pressure remain constant while volume expands? Since II is the isothermal, for any path that is above II, what must happen to the temperature?

AM

5. Jan 1, 2007

### twotaileddemon

Would these be acceptable responses now?

d. In which process or processes is heat removed from the gas? Explain.
∆U = Q – W.
By definition, adiabatic processes have no heat lost or gained, resulting in Q = 0. Therefore, any process above the adiabatic one will have a positive internal energy and any process below the adiabatic process will have a negative internal energy. Therefore, in process IV there will be a negative internal energy and so heat is removed from the gas.

e. In which process or processes does the root-mean-square speed of the gas molecules increase? Explain.
Vrms = sqrt (3RT/Mm)
If the temperature decreases, the speed of the gas particles decreases as well. An isothermal process has a constant temperature; therefore the process above it, process I, must have an increase in temperature. As such, when the temperature increases, the root-mean-square speed of the gas molecules must increase as well.

6. Jan 1, 2007

### twotaileddemon

Oh, and thanks for the help ^^; I appreciate it honestly. =)

7. Jan 2, 2007

### Andrew Mason

It is not a question of having positive or negative (change) in internal energy. It is whether the heat flow is negative or positive. If there is a loss of internal energy but work is done, there may be no loss of heat - as in the adiabatic process.

If dQ is negative, heat is removed. Since dQ = dU + dW, it is not simply a matter of looking at dU. In this case, the curve for dQ = 0 (or dU + dW = 0) is III. If dU + dW > 0 (PdV > - dU where dV > 0) then there is a loss of internal energy that is not as great as the positive work done. This means that heat flow is into the gas. If dU + dW < 0, there is a loss of internal energy that is greater than the positive work done. This means that heat flow is out of the gas. If the PV graph is below the adiabatic (III) then the loss of internal energy is greater than in the adiabatic case AND the area below the graph (work done by the gas) is less than the adiabatic. So there must be heat flow out of the gas.
This is just a matter of PV = nRT. If at all points in the process, PV is greater than PV on the isotherm T0, then T must have increased. If T > T0 then the internal energy (dU = nCvdT) must be greater.

AM

8. Jan 2, 2007

### twotaileddemon

So is it correct in saying that, for part d, heat is removed from process IV because there is less internal energy and heat is removed from process I and II because the area under the curve is smaller and there is less work?

9. Jan 2, 2007

### Andrew Mason

For d, in process IV heat is removed from the gas because there is no work done (area under IV is 0) and the temperature is lower (PV is reduced so nRT is reduced).

I don't understand why you think the area under the I and II is smaller (than what?). Heat is added to the gas in I and II. In process I, PV is increased so temperature is increased (internal energy increased) AND work is done (which would use energy), so energy (heat) must have been added to the gas. In II, the temperature is the same but work is done by the gas. Since the internal energy is the same (T is the same) AND work is done, energy (heat) must have been added.

AM

10. Jan 2, 2007

### twotaileddemon

Oh.. I think I figured out what I was confused. I assumed I stopped at the next process, figuring the work done for process I would be only that half of the top part above process II.. okay thanks ^_^;;
This helped alot, especially since I have a midterm next week and most likely needed to know that!