- #1

asdf1

- 734

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Can someone explain why U=Cv(dT) and Enthalphy=Cp(dt) for all processes in thermodynamics?

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- Thread starter asdf1
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- #1

asdf1

- 734

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Can someone explain why U=Cv(dT) and Enthalphy=Cp(dt) for all processes in thermodynamics?

- #2

pocoman

- 10

- 0

Enthalpy W is defined : W = U +PV => dW = dU +PdV +VdP = dU + dA + VdP = dQ + VdP, in isobaric processes dP=0 so dW=dQ (while P=const) =Cp*dT.

- #3

LeonhardEuler

Gold Member

- 860

- 1

[tex]dU=(\frac{\partial U} {\partial T})_VdT+(\frac{\partial U} {\partial V})_TdV[/tex]

which, by definition of [itex]c_V[/itex] is

[tex]dU=c_VdT+(\frac{\partial U} {\partial V})_TdV[/tex]

So, dU is not always equal to [itex]c_VdT[/itex]. However, for a perfect gas [tex](\frac{\partial U} {\partial V})_T[/tex] is always equal to zero, so that means that [itex]dU=c_Vdt[/itex] for all processes involving perfect gases. That is probably what you meant. The dH equation is simmilar.

It seems strange at first that you would use the constant volume heat capacity for any process, including one at constant pressure. But this really isn't so strange. Fitrst of all, using [itex]c_P[/itex] would make no sense to find [itex]\Delta U[/itex] because this is defined as [itex](\frac{\partial H} {\partial T})_P[/itex] and notice that dU doesn't even appear there. Secondly, supposing the process is at constant pressure: U is a state function so we can imagine any path between the beginning and end points we want to compute [itex]\Delta U[/itex]. Imagine the gas is first heated at constant volume. [itex]\Delta U[/itex] for this process is clearly [itex]c_V\Delta T[/itex]. Now let the gas expand at constant volume back to its original pressure. [itex]\Delta U[/itex] for this process is [itex](\frac{\partial U} {\partial V})_T \Delta V[/itex]. But [itex](\frac{\partial U} {\partial V})_T[/itex] is zero for a perfect gas, so the total [itex]\Delta U[/itex] is [itex] c_V \Delta T[/itex]

- #4

asdf1

- 734

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wow! thanks! :)

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