# Heat Capacity and First Thermodynamic Law

## Main Question or Discussion Point

I have a question about the deduction of First Thermodynamic Law. The book that I have is written by Paul A. Tipler and Gene Mosca and it is called Physics: For Scientists and Engineers.

The way to deduct it is given here:

@constant volume,
Qv=Cv(T2-T1)
Because W=0, E2-E1=Cv(T2-T1)
After all, Cv=d(E2-E1)/d(T2-T1)

@constant pressure,
Given that E2-E1=Qp-P(V2-V1) and Qp=Cp(T2-T1)
We have Cp(T2-T1)=(E2-E1)+P(V2-V1)
Now, the author replaces E2-E1 with Cv(T2-T1), I cannot understand this because he incites something under constant volume into a formula under constant pressure.

My question is what determines the internal heat in a system is defined by CvP, which can even be used in a situation in which a different condition is given.

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Mapes
Homework Helper
Gold Member
Hi FriedrichLuo, welcome to PF. Yours is a very common question, as it is disconcerting to see a "constant-volume" parameter being used in a constant-pressure process. But energy is a state variable (its value is process independent), and for this system the relationship E2-E1 = Cv(T2-T1) holds for all processes.

Hi FriedrichLuo, welcome to PF. Yours is a very common question, as it is disconcerting to see a "constant-volume" parameter being used in a constant-pressure process. But energy is a state variable (its value is process independent), and for this system the relationship E2-E1 = Cv(T2-T1) holds for all processes.

dU=dQ+dW, dU is a state parameter and dQ and dW are process parameters. Therefore, dQ(or Cv(T2-T1) holds for any situations where process is involved. And dU is a result when a new state is reached. In other words, dQ and dW are what is really happening; dU is only concept to show the resultant of two true processes.

I hope I get it right. :D

Mapes
Homework Helper
Gold Member
This sounds like a fine way of thinking about things, but make sure you don't mix your differential and finite values. You can say

$$\delta Q=C_V\,dT$$

or

$$Q=C_V(T_2-T_1)$$

but not mix them.