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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?

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  • $\begingroup$ 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. $\endgroup$
    – Steeven
    Commented Oct 11, 2015 at 13:09

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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.

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  • $\begingroup$ We get Cv when total heat supplied to the system results solely in increasing the internal energy of the system by keeping volume constant. And CvdT term represents internal energy of the system which is equal to total heat added since work done is zero because change in volume is zero. So how is there a compression work if volume is constant. $\endgroup$
    – undertrial
    Commented Oct 11, 2015 at 14:09
  • $\begingroup$ There isn't compression work if volume is constant. But the 1st law of thermodynamics wouldn't be complete without this term, which also covers the case where there is compression work. $\endgroup$
    – David White
    Commented Oct 11, 2015 at 17:21
  • $\begingroup$ In a situation where compression takes place, the CvdT term should not be used in 1st law of thermodynamics equation because we get this Cvdt term when volume is assumed to be constant. The terms used in equation seems contradictory i.e. Cv and pdV. One is for constant volume other is for change in volume so either one of them should satisfy the equation. $\endgroup$
    – undertrial
    Commented Oct 12, 2015 at 17:12
  • $\begingroup$ 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 $\delta Q$. $\endgroup$
    – David White
    Commented Oct 12, 2015 at 21:35

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