Relation between enthalpy and Cv

**MY ATTEMPT**


We already know that $\Delta U=nC\_V\,\Delta T$
which is for constant volume


Now we know that $\Delta H = \Delta U + p\,\Delta V$


so $\Delta H = nC\_V\,\Delta T + 0$ (as at constant volume $\Delta V = 0$)


$$\Delta H=nC\_V\,\Delta T$$


But I have never such an equation anywhere.


Please verify.




For an ideal gas, your equation for $\Delta U$ applies not only to constant volume processes, but to all processes. It's just that, in experiments to measure Cv, it is very convenient to use a constant volume path because, under those circumstances, no work is done, in which case $\Delta U$ is just equal to the amount of heat added. So this gives you a direct measurement of Cv.


As far as $\Delta H$ is concerned, the correct equation is $\Delta H=\Delta U+\Delta (PV)$. But, from the ideal gas law, $\Delta (PV)=nR\Delta T$. So what does that give you for $\Delta H$?


 2