Thermo - double Vrms, find heat required

[undagada]

Homework Statement

n moles of a diatomic gas with C_V =5/2 R has initial pressure p_i and volume V_i. 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, p_i , and V_i

Homework Equations

PV = 2/3 N ε_avg
ε_avg = 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:
ΔE = \frac{5}{2} nRΔT
= \frac{5}{2} nR(\frac{4P_iV_i}{nR} - \frac{P_iV_i}{nR})
= \frac{15}{2} P_iV_i

However, the answer in the back is:

\frac{15n+3}{2} P_iV_i

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

 

[TSny]

Welcome to PF!

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:
ΔE = \frac{5}{2} nRΔT
= \frac{5}{2} nR(\frac{4P_iV_i}{nR} - \frac{P_iV_i}{nR})
= \frac{15}{2} P_iV_i

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:
\frac{15n+3}{2} P_iV_i

I think there must be a misprint in this expression. Q should be proportional to the number of moles.

 

[undagada]

This would cause PV to increase by a factor of 16.

You're right, it should be 2 Pi and 2 Vi.

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.

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

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!