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

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SUMMARY

The discussion focuses on calculating the specific heat at constant volume (Cv) for an ideal gas undergoing an adiabatic process. The initial and final temperatures are given as 273K and 433K, respectively, with the gas compressed to half its original volume. The key equations utilized include Cv = dU/dT, dU = dq + dw, and the relationship between pressure and volume in adiabatic processes. The final expression derived for Cv is Cv = -nR*ln(1/2), confirming the calculations based on the principles of thermodynamics.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law (PV = nRT)
  • Knowledge of specific heat capacities (Cv and Cp)
  • Ability to apply calculus in thermodynamic equations
NEXT STEPS
  • Learn about the derivation of the adiabatic process equations
  • Study the relationship between Cp and Cv, specifically Cp - Cv = R
  • Explore the concept of the heat capacity ratio (γ) and its significance
  • Investigate the implications of temperature independence of Cv in real gases
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those tackling problems related to ideal gases and adiabatic processes. It is also useful for educators and professionals in physics and engineering fields who require a solid understanding of heat transfer concepts.

Ghodsi
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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?)
 
Last edited:
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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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