# Thermal Processes.

1. Apr 23, 2005

### dalitwil

Ok guys, my last questions of the semester, Yay!

So I have two of them, one pertaining to a figure (attached), which has me terribly confused:

1.) In the figure (attached), if P1 = 133 kPa, V1 = 1123 cm3 and P2 = 200 kPa, V2 = 10000 cm3, what is the heat absorbed (+) or liberated (-), to the nearest joule, in Path 2? Assume the change in internal energy is 8900 J.

My Work (Reasoning):
I am assuming that W=0, simply because the path appears to be only vertical, So using ΔU=Q-W, we can say that ΔU=Q (because W=0) and ΔU is given as 8900 J, so wouldn't this be the answer? (It is not)

2.) 10 moles of a gas in thermal contact with an oil bath at temperature 300 K is compressed isothermally from a volume of 7541 cm3 to a volume of 1561 cm3. To the nearest joule what is the work done by the piston?

My Work:
W would simply equal nRT*ln(Vf/Vi)
Plugging in the knowns leaves me with -131
This is incorrect. (using the reasoning that ΔU=0)

Any ideas guys?

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2. Apr 23, 2005

### OlderDan

I just took a quick look- so this is totally off the cuff, but have you included the whole path? I think the intent is to go from point 1 to point 2, part horizontal and part vertical

3. Apr 23, 2005

### Andrew Mason

As Dan says, the path is from 1 to 2 so you have to look at the horizontal part. I am a little confused by the assumption that change in internal energy is 8900 J.

Your approach is correct. Try the numbers again. I think you forgot to factor in the temperature.[/QUOTE]

AM

4. Apr 24, 2005

### dalitwil

The first question is extrememly confusing, because it must mean that the temperature is being held constant (and also not given) so the change in internal energy should equal 0, which it doesn't. I cannot find W, without temperature and moles. I am really stuck

5. Apr 24, 2005

### Andrew Mason

The temperature is not constant:

$$T = PV/nR$$

So:$$T_2/T_1 = P_2V_2/P_1V_1 \ne 1$$

I think you have to ignore the assumption that the change in internal energy is 8900 J. The change in internal energy is:

$$\Delta U = \Delta (PV) = P_2V_2 - P_1V_1 = 1850 J$$

The work done by the gas from 1 to 2 is $W = P_1(V_2 - V_1)$

The heat added is the sum of the work done + change in internal energy.

AM