# Which one is correct $\Delta U = nC_v \Delta T$ or $q = nC_v \Delta T$?

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?

• 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. Commented Apr 30, 2020 at 16:40
• done! now could you plz answer the question. Commented Apr 30, 2020 at 16:45

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.

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 • 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 Commented Apr 30, 2020 at 16:35 • 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. Commented Apr 30, 2020 at 16:38 • No, the value of$\Delta U$will not be equal to Q. Commented Apr 30, 2020 at 17:10 • 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\$ Commented Apr 30, 2020 at 17:12
• 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 Commented Apr 30, 2020 at 17:25