How to calculate Cv


by Ghodsi
Tags: heat capacity
Ghodsi
Ghodsi is offline
#1
Feb9-12, 08:44 PM
P: 4
1. The problem statement, all variables and given/known data

When one mole of an ideal gas is compressed adiabatically to one-half of its original volume, the temperature of the gas increases from 273 to 433K. Assuming that Cv is independent of temperature, calculate the value of Cv for this gas.


2. Relevant equations

Cv = dU/dT
dU = dq + dw
dq = 0 for adiabatic processes, thus dU=dw
PV = nRT


3. The attempt at a solution

Cv = -pdV / dT
Cv = (-nRT/V)(dV/dT)

I'm stuck here.
Assuming I'm correct thus far, do I use the initial or final values for T and V (i.e. do I use 273K or 433K?)
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beet
beet is offline
#2
Feb9-12, 10:26 PM
P: 2
I think we might be in the same class...

I've been trying to verify my solution, but no luck so far. This is what I got:

Cv = (dU/dT)
dU = dq + dw, but dq = 0, so dU = dw and Cv = dU/dT

w = -nRTln(V2/V1), but V2 = 1/2V1, so w = -nRTln(1/2), and dw = -nR*ln(1/2)*dT

Substitute the last equation for dw in Cv=dw/dT and you get Cv = -(nR*ln(1/2)*dT)/dT which simplifies to Cv = -nR*ln(1/2).

That's what I got, but I'm not confident that it's correct.
Ghodsi
Ghodsi is offline
#3
Feb9-12, 10:48 PM
P: 4
Elber 10am MWF?
Quote Quote by beet View Post
I think we might be in the same class...

I've been trying to verify my solution, but no luck so far. This is what I got:

Cv = (dU/dT)
dU = dq + dw, but dq = 0, so dU = dw and Cv = dU/dT

w = -nRTln(V2/V1), but V2 = 1/2V1, so w = -nRTln(1/2), and dw = -nR*ln(1/2)*dT

Substitute the last equation for dw in Cv=dw/dT and you get Cv = -(nR*ln(1/2)*dT)/dT which simplifies to Cv = -nR*ln(1/2).

That's what I got, but I'm not confident that it's correct.

beet
beet is offline
#4
Feb9-12, 10:56 PM
P: 2

How to calculate Cv


Yeah.
Meemo
Meemo is offline
#5
Feb9-12, 11:11 PM
P: 4
Quote Quote by Ghodsi View Post
1. The problem statement, all variables and given/known data

When one mole of an ideal gas is compressed adiabatically to one-half of its original volume, the temperature of the gas increases from 273 to 433K. Assuming that Cv is independent of temperature, calculate the value of Cv for this gas.


2. Relevant equations

Cv = dU/dT
dU = dq + dw
dq = 0 for adiabatic processes, thus dU=dw
PV = nRT


3. The attempt at a solution

Cv = -pdV / dT
Cv = (-nRT/V)(dV/dT)

I'm stuck here.
Assuming I'm correct thus far, do I use the initial or final values for T and V (i.e. do I use 273K or 433K?)
I think you should use
T1/T2 = (V2/V1)^γ-1
then you also find the value of
P1 and P2 from
P1V1^γ= P2V2^γ

The put the values in adiabatic process equation
∂W = (P1V1-P2V2)/γ-1
Then use your formulae
Ghodsi
Ghodsi is offline
#6
Feb10-12, 12:53 AM
P: 4
Thanks guys. This is pretty crucial assistance.
dpsguy
dpsguy is offline
#7
Feb10-12, 03:37 AM
P: 69
Quote Quote by Meemo View Post
I think you should use
T1/T2 = (V2/V1)^γ-1
Find γ from the above. Then use:

Cp-Cv = R (gas constant)
(Cp/Cv) = γ

Eliminate Cp from these two equations to get Cv.


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