# 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. – David White Apr 30 '20 at 16:40
• done! now could you plz answer the question. – Priyanuj Bora Apr 30 '20 at 16:45
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$$ you defined in the first case should be valid here too, hence generalising the expression • I know that! but plz refer to physics.stackexchange.com/questions/336945/…. here the first answer (with the most votes) reads out that$\Delta U = nC_v \Delta T$can be used as long as the gas is ideal. Even in many questions, I found answers where the relation,$\Delta U = nC_v \Delta T$was used even tough the volume wasn't constant. Please elaborate more – Priyanuj Bora Apr 30 '20 at 16:35 • What I have interpreted from your answer is that only in the cases where volume is constant we can use the relation,$\Delta U = nC_v \Delta T$. But in many problems, where the volume isn't constant, the above relation is used. – Priyanuj Bora Apr 30 '20 at 16:38 • No, the value of$\Delta U$will not be equal to Q. – Elendil Apr 30 '20 at 17:10 • But, if the change in temperature is same, irrespective of the change in volume,$\Delta U$will be the same. Q does not affect the value of$\Delta U\$ – Elendil Apr 30 '20 at 17:12
The precise definition of Cv involves the partial derivative of U with respect to temperature at constant volume: $$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.