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In many books, I've found that MOLAR HEAT CAPACITY $(C)$ is defined as the amount of heat required to change the temperature of 1 mole of a substance by 1K, which mathematically translates to $C=\frac{q}{n\Delta T}$ and at constant volume, it becomes $C_v=\frac{q}{n\Delta T}$.

But in some examples, I've also seen $C_v=\frac{\Delta U}{n\Delta T}$. Even in many questions, I found answers where the relation, $\Delta U=nC_v\Delta T$ was used even though the volume wasn't constant.

So which one is correct? does that mean heat $q=C$ at all times?

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    $\begingroup$ There is also a constant pressure specific heat, so it is standard notation to place either a subscript "p" or a subscript "v" after the "C" to indicate which specific heat you are talking about. $\endgroup$ – David White Apr 30 '20 at 16:40
  • $\begingroup$ done! now could you plz answer the question. $\endgroup$ – Priyanuj Bora Apr 30 '20 at 16:45
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    $\begingroup$ Krishna already answered your question. $\endgroup$ – David White Apr 30 '20 at 16:53
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BOTH. In an isochoric process, the gas does not do any work - since $W=P\Delta V$ and the volume, being constant renders the work done zero.

So,the first law of thermodynamics becomes, $$Q=\Delta U+W$$ Since $W=0$, $$Q=\Delta U$$ So, both the formulae are correct.

EDIT The relation $Cv=\frac{\Delta U}{n\Delta T}$ is always applicable. The relation between internal energy and temperature, as you have read from the answer, can be derived from kinetic theory (using the equipartition theorem). $$U=\frac f2 nrt$$ Differentiating this, you will get the expression you seek. So, the term $C_v$ is in the equation only because it fits the necessary condition (it is a lot more easier to measure than the degrees of freedom).

So, in short, the $C_v$ has to be measured under constant volume, because, under constant volume, $C_v=\frac f2 R$$. But, once you have found that out, you can expect it to valid everywhere.

For an even more intuitive example, I'll give you an example.

Consider, a a system (gas) at a temperature $T_1$. Now, you give it some energy (say, under constant volume)and take its temperature to $T_2$. Let the change in internal energy be $\Delta U$. Let this be case one.

In the next case, you supply more energy, but still, bring the temperature down by the same amount (i.e. to $T_2$). Now, since the internal energy must depict the temperature (to some extent) of the gas, the change in internal energy must remain the same. So, the $\Delta U$ you defined in the first case should be valid here too, hence generalising the expression

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  • $\begingroup$ I know that! but plz refer to physics.stackexchange.com/questions/336945/…. here the first answer (with the most votes) reads out that $\Delta U = nC_v \Delta T$ can be used as long as the gas is ideal. Even in many questions, I found answers where the relation, $\Delta U = nC_v \Delta T$ was used even tough the volume wasn't constant. Please elaborate more $\endgroup$ – Priyanuj Bora Apr 30 '20 at 16:35
  • $\begingroup$ What I have interpreted from your answer is that only in the cases where volume is constant we can use the relation, $\Delta U = nC_v \Delta T$. But in many problems, where the volume isn't constant, the above relation is used. $\endgroup$ – Priyanuj Bora Apr 30 '20 at 16:38
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    $\begingroup$ No, the value of $\Delta U$ will not be equal to Q. $\endgroup$ – Elendil Apr 30 '20 at 17:10
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    $\begingroup$ But, if the change in temperature is same, irrespective of the change in volume, $\Delta U$ will be the same. Q does not affect the value of $\Delta U$ $\endgroup$ – Elendil Apr 30 '20 at 17:12
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    $\begingroup$ Of course, yes, as long as the gas does not do work. Its fairly intuitive. You compress a gas temperature increases. You give it heat, temperature increases. If you do both, then the increase will be more than them taken independently $\endgroup$ – Elendil Apr 30 '20 at 17:25
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The precise definition of Cv involves the partial derivative of U with respect to temperature at constant volume: $$C_v=\frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_V$$For an ideal gas, U and $C_v$ are functions only of temperature, so it doesn't matter if the volume changes. But, for other equations of state, this is not necessarily the case.

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