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In many places I have come across the following derivation

Work done in isothermal expansion

By first law of thermodynamics

  1. $\mathrm dU=Q+W$

  2. As $\mathrm dU=nC_V\,\mathrm dT$, and in isothermal process $\mathrm dT = 0$

$\therefore\mathrm dU=0$

$W=-Q$
i.e heat absorbed is completely converted into work

The second step doesn't make any sense at all. As far as my understanding $\mathrm dU=nC_V\,\mathrm dT$ is true only when volume is constant, but it's clearly mentioned expansion in the statement.

So how to justify the use of $\mathrm dU=nC_V\,\mathrm dT$ expression even when volume is not constant?

Please consider the fact that I'm a school student so please answer in a simple manner.

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5 Answers 5

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The reason why statement #2 doesn't make sense to you is because it is not generally true!

For a closed system of constant composition, with P-V work only,

$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$

The difference between that statement and your statement is the extra term at the end, which is the amount by which U changes with V at constant T, times the change in V.

[Minor note: I've dropped the "n" you use, as I'm using total heat capacities, not molar heat capacities.]

However, if we have an ideal gas, U is independent of V, which means:

$$\left(\frac{\partial U}{\partial V}\right)_T=0,$$

Hence, with an ideal gas, the term on the right goes away, giving us your statement #2:

$$dU = C_v dT$$

I.e., your statement #2 doesn't follow directly from #1. Instead, it's specific to ideal gases.

So: If a process invoving an ideal gas is carried out at constant pressure instead of constant volume, it is true that $dV$ may not be zero. But since that non-zero $dV$ doesn't matter when it comes to calculating $dU$ for ideal gases, we can use $C_V dT$ even when $dV \neq 0$.

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  • $\begingroup$ May I know how we got this relation $$\left(\frac{\partial U}{\partial T}\right)_V dT = C_v dT$$ , I’m aware that it’s a general fact that heat added at constant volume is Equal to increase in internal energy. Is it possible to prove this relation mathematically $\endgroup$
    – Chemist
    Commented Jan 12, 2020 at 2:12
  • $\begingroup$ @Chemist Any standard physical chemistry textbook will cover this. For instance, go to this URL at Google Books for Castellan, Physical Chemistry, 3rd Ed., and click on the link for Page 117: books.google.com/… $\endgroup$
    – theorist
    Commented Jan 14, 2020 at 5:55
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I am a 12th grader, so I'll explain in the most colloquial way possible.

  1. Internal energy is a state function. A state function is a property whose value does not depend on the path taken to reach that specific function or value. ... Path functions are functions that depend on the path taken to reach that specific value. (1)

  2. This implies a derivation for internal energy will stand true, even if if you change the process, as far as you keep the initial and final conditions same, as it is a state function as defined above, so it will depend only on the initial and final conditions and not the path taken to reach so (By conditions I mean the conditions that internal energy depends on, which is merely temperature)

  3. The significance of this is that $C_v$ is merely appearing because you derived it using an isochoric process. If you would have used an Isobaric process, you would see ($C_p$-R), now would you wonder why $C_p$ is appearing when the pressure is not constant?

So basically $C_v$ appears as a consequence of the first law of thermodynamics. Not as a consequence of constant volume

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  • $\begingroup$ Thank you for answering , Moreover as mentioned in 2nd statement why is it essential to keep the initial and final conditions same what are its implications, kindly explain (12th graders are undergrads) $\endgroup$
    – Chemist
    Commented Jan 1, 2020 at 17:05
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Zenix
    Commented Jan 1, 2020 at 17:10
  • $\begingroup$ I have added it in my answer check it out $\endgroup$ Commented Jan 1, 2020 at 17:17
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This is related to the difference between how Cv is measured experimentally and how it is applied in solving practical problems. To measure Cv experimentally, we hold the volume constant and determine the change in internal energy U by measuring the amount of heat added. However, for practical purposes, we know that, for an ideal gas, U and Cv are independent of volume (and depend only on temperature). So, once Cv is known from a constant volume experiment, we can use the measured value to calculate the change in internal energy for any situation involving a change in temperature (and volume). For a situation where the temperature does not change, the change in U is zero, irrespective of what the volume is doing.

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This has to do with the simplifications we make regarding the energy of molecules in an ideal gas (they've been largely discussed here if you want to have a look). But essentially, in an ideal monoatomic gas, we can state that all the energy is translational kinetic energy, and so we can write that the mean energy is: $$U=n \frac{3}{2}RT=n c_vT$$

So, the expression $\delta U = nc_v \delta T$ is only valid for ideal gases.

In all other cases: $\delta U = T\delta S - p\delta V$, where $S$ is the entropy of the system.

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You may calculate dU and/or dH in any transformation, at constant volume or constant pressure, because U and H are state functions. These variations of U and H may not be easy to obtain, but they can always be calculated.

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