Thermo - double Vrms, find heat required

In summary, the conversation discussed a problem involving a diatomic gas undergoing a process in which the pressure is directly proportional to the volume until the rms speed of the molecules has doubled. The solution involved calculating the change in internal energy and accounting for any work done by the gas to find the amount of heat required for the process. There may have been a misprint in the answer provided in the textbook.
  • #1
undagada
2
0

Homework Statement


n moles of a diatomic gas with CV =5/2 R has initial pressure pi and volume Vi. The gas undergoes a process in which the pressure is directly proportional to the volume until the rms speed of the molecules has doubled.
a. Show this process on a pV diagram.
b. How much heat does this process require? Give your answer in terms of n, pi , and Vi

Homework Equations


PV = 2/3 N εavg
εavg = 1/2 mv2
PV = nRT
E = 5/2 nRT
Q = ΔE

The Attempt at a Solution


Part a was done simply by linearly increasing Pi and Vi to 4 Pi and 4 Vi respectively.

Part b:
[itex]ΔE = \frac{5}{2} nRΔT[/itex]
[itex] = \frac{5}{2} nR(\frac{4P_iV_i}{nR} - \frac{P_iV_i}{nR})[/itex]
[itex] = \frac{15}{2} P_iV_i[/itex]

However, the answer in the back is:

[itex]\frac{15n+3}{2} P_iV_i[/itex]

Not sure where the n and +3 came from, maybe PV = 2/3 N εavg? Plugging that in doesn't seem to work though.
 
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  • #2
Welcome to PF!

undagada said:

Homework Equations


...
Q = ΔE
Does this equation account for any work done by the gas?

The Attempt at a Solution


Part a was done simply by linearly increasing Pi and Vi to 4 Pi and 4 Vi respectively.
This would cause PV to increase by a factor of 16.

Part b:
[itex]ΔE = \frac{5}{2} nRΔT[/itex]
[itex] = \frac{5}{2} nR(\frac{4P_iV_i}{nR} - \frac{P_iV_i}{nR})[/itex]
[itex] = \frac{15}{2} P_iV_i[/itex]
OK, now you have PV increasing by a factor of 4. I believe this is correct for ΔE. However, to get Q you will need to take into account any work done by the gas.

However, the answer in the back is:
[itex]\frac{15n+3}{2} P_iV_i[/itex]
I think there must be a misprint in this expression. Q should be proportional to the number of moles.
 
  • #3
TSny said:
This would cause PV to increase by a factor of 16.
You're right, it should be 2 Pi and 2 Vi.
TSny said:
OK, now you have PV increasing by a factor of 4. I believe this is correct for ΔE. However, to get Q you will need to take into account any work done by the gas.
Okey dokey.

[itex]\Delta E = \frac{15}{2} P_iV_i[/itex]
[itex] W = \int pV [/itex]
[itex]\Delta E = Q -W_s[/itex]
[itex] \frac{15}{2} P_iV_i = Q - ( \frac{1}{2} (2V_i - V_i)(2P_i - P_i) + (P_i V_i) ) [/itex]
[itex] Q = \frac{15}{2} P_iV_i + \frac{P_i V_i}{2} + P_i V_i [/itex]
[itex] Q = \frac{18}{2} P_iV_i[/itex]

That looks a little better. I would not be surprised if the answer is a misprint; that has happened a lot for this textbook. (Knight Physics 3rd ed.)

Thanks for your help!
 
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