Specific heat at constant volume

  • Thread starter Bipolarity
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  • #1
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[tex] C_{V} = \frac{∂U}{∂T} [/tex]

This is the specific heat at constant volume so I assume it can only be used at constant volume. However, my textbook uses this to derive the following equation for reversible adiabatic expansion:

[tex] P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ} [/tex]

Why are we allowed to use [itex]C_{V}[/itex] when it only works in isovolumetric processes?

BiP
 

Answers and Replies

  • #2
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How is Cv used to derive the equation for adiabatic transformation?
Can you show it here?
 
  • #4
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The change in internal energy has the same expression for any process between two states. For ideal gas is
[tex]\Delta U = nC_v\Delta T[/tex]
The amount of heat is dependent on the type of process. It is [tex]Q = nC_v\Delta T[/tex]
only for constant volume process.
 
  • #5
775
2
The change in internal energy has the same expression for any process between two states. For ideal gas is
[tex]\Delta U = nC_v\Delta T[/tex]
The amount of heat is dependent on the type of process. It is [tex]Q = nC_v\Delta T[/tex]
only for constant volume process.

Superb! Thanks!

BiP
 
  • #6
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[tex] C_{V} = \frac{∂U}{∂T} [/tex]

This is the specific heat at constant volume so I assume it can only be used at constant volume. However, my textbook uses this to derive the following equation for reversible adiabatic expansion:

[tex] P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ} [/tex]

Why are we allowed to use [itex]C_{V}[/itex] when it only works in isovolumetric processes?

BiP

For an ideal gas, the internal energy is a function only of temperature, such that dU = CvdT always. For an adiabatic expansion, dQ = 0, so that

dU = CvdT = -pdV
 

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