# Why is change in volume ($dV$) is not taken as zero when we use heat capacity at constant volume ($C_v$) in first law of thermodynamics equation

Why is change in volume ($dV$) is not taken as zero when we use heat capacity at constant volume ($C_v$) in first law of thermodynamics equation.

$$dQ = C_v dT + pdV$$ Here $$C_v = (dQ/dT) = \text{constant}$$ since $dQ = dU$ (since $dV = 0$).
Therefore $$C_v = (dU/dT) = constant$$

So why in first equation, $dV$ is not taken as zero?

• The $pdV$ term is simply adding what is missing. That means, the $C_vdT$ term gives the energy added for a constant volume situation. The $pdV$ term then adds the rest to end with the correct amount of energy. Oct 11, 2015 at 13:09

Your first equation is just saying that the differential heat added to a system is equal to the differential heat from the $C_v dT$ term + the differential heat added to the system via compression work. In other words, the total differential heat added to a system is equal to the linear combination of constant-volume specific heat and compression work.
• YOU have to decide what remains constant, based on your particular process (e.g., isothermal, adiabatic, etc.), and YOU have to decide which term equates to zero, and YOU have to drop that term from the equation. Which terms get dropped depends on the particular process circumstances that you are dealing with. In addition, I wouldn't normally bastardize the first law, as you did in the top equation, by equating $\delta U$ to $\delta Q$. Oct 12, 2015 at 21:35