# Thermo - double Vrms, find heat required

1. Feb 23, 2016

1. The problem statement, all variables and given/known data
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

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

3. 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 εavg? Plugging that in doesn't seem to work though.

2. Feb 23, 2016

### TSny

Welcome to PF!

Does this equation account for any work done by the gas?

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

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.

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

3. Feb 23, 2016

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