As pressure is increased, do we require more energy to increase the temperature from given temperature for the same mass. For example, if we heat water at 1 atm to raise its temperature by 20 degree Celsius, will the heat required be the same when pressure is 3 atm?
[Physics] Does specific heat change with pressure? If so, why
thermodynamics
Best Answer
I shall try to explain using first law of thermodynamics and Maxwell' relations. For simplicity I am considering $C_v$ . You may extend the argument to $C_p$ as well. Let us consider the change in $C_v$ with w.r.t $V$. From the definition of $C_v$ we can write,
$\left(\frac{dC_v}{dV}\right)_T = \left[\frac{d}{dV}\left(\frac{dU}{dT}\right)_V\right]_T ...(1)$
Switch order of differentiation and from the first law of thermodynamics we can write
$\left[\frac{d}{dT}\left(\frac{dU}{dV}\right)_T\right]_V = \left[\frac{d}{dT}\left(T \frac{dS}{dV} - P \right)_T\right]_V .. (2)$
From Maxwell's relation we have
$ \left(\frac{dS}{dV}\right)_T = \left(\frac{dP}{dT}\right)_V .. (3) $
Substitute Eqn $(3)$ in $(2)$ and after applying product rule, you may obtain
$\left(\frac{dC_v}{dV}\right)_T = T \left(\frac{d^2P}{dT^2}\right)_V$
For an ideal gas or a gas having an equation of state of the form $PV = kT$ or a linear function of $T$, the second derivative of $P$ w.r.t $T$ vanishes and therefore no dependence. For any other equation of state, this is the dependence of specific heat w.r.t to pressure or volume. You may work out in a similar fashion for $C_p$ w.r.t $P$. Only difference is that the thermodynamic quantity will be Enthalpy and not Internal Energy