Calc Cv from T & V: How to Calculate Cv Homework

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The discussion revolves around calculating the specific heat at constant volume (Cv) for an ideal gas compressed adiabatically from an initial temperature of 273K to a final temperature of 433K and halved volume. Participants are uncertain whether to use initial or final values for temperature and volume in their calculations. The equations discussed include Cv = dU/dT and the work done during the process, leading to Cv = -nRln(1/2). Suggestions are made to utilize the relationship between temperature and volume ratios to find the specific heat ratio (γ) and subsequently derive Cv. The conversation highlights the importance of correctly applying thermodynamic principles to solve the problem.
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Homework Statement



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.

Homework Equations



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

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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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.
 
Elber 10am MWF?
beet said:
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.
 
Yeah.
 
Ghodsi said:

Homework Statement



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.


Homework Equations



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


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
 
Thanks guys. This is pretty crucial assistance.
 
Meemo said:
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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