Question

The molar heat capacity of a process as $C = {C_V} + aV$, where a is a constant. Find the equation of the process in the V-T variable.

Answer 1

Hint: The molar heat capacity of a process is the amount of the heat added to a substance to change its temperature by one unit. In order to find the relation between V-T for the process, we need to establish a relation between the given equation and the standard equation of molar heat capacity. Then after making certain changes we’ll have the required equation.

Formula used:
$\eqalign{
  & C = {C_V} + P\dfrac{{dV}}{{dT}} \cr
  & PV = RT \cr} $

Complete answer:
Generally, molar heat capacity is the amount of heat required to increase the temperature by one unit for one mole of a substance.
For a thermodynamic process, it is given by
$C = {C_V} + P\dfrac{{dV}}{{dT}}$
Where,
C is the heat capacity
Cv is the heat capacity at constant volume
P is the pressure
V is the volume
T is the temperature
We’ve already been given an equation for the molar heat capacity. Equating these both, we have
$\eqalign{
  & C = {C_V} + P\dfrac{{dV}}{{dT}} \cr
  & \Rightarrow {C_V} + aV = {C_V} + P\dfrac{{dV}}{{dT}} \cr
  & \Rightarrow aV = P\dfrac{{dV}}{{dT}} \cr
  & \Rightarrow \dfrac{{dV}}{V} = \dfrac{a}{P}dT \cr} $
We already know the ideal gas equation $PV = RT$. From this, we have P as $P = \dfrac{{RT}}{V}$. Substituting this in the former equation we get
$\eqalign{
  & \dfrac{{dV}}{V} = \dfrac{a}{P}dT \cr
  & \Rightarrow \dfrac{{dV}}{V} = a \times \dfrac{V}{{RT}}dT \cr
  & \Rightarrow \dfrac{{dV}}{{{V^2}}} = \dfrac{a}{R}\dfrac{{dT}}{T} \cr} $
Integrating this equation on both sides, we’ll have
$\eqalign{
  & \int {\dfrac{{dV}}{{{V^2}}}} = \dfrac{a}{R}\int {\dfrac{{dT}}{T}} \cr
  & \Rightarrow - V = \dfrac{a}{R}\ln T \cr
  & \Rightarrow \ln T = - \dfrac{R}{a}V \cr
  & \Rightarrow T = {e^{ - \dfrac{R}{a}V}} \cr} $
Therefore, the equation of the given thermodynamic process in terms of V-T is $T = {e^{ - \dfrac{R}{a}V}}$.

Note:
The molar heat capacity of a substance can be a little hard to remember. In that case, you can simply derive it from the first law of Thermodynamics, which is mathematically given by $dQ = dU + PdV$. Dividing this equation with $dT$ on both the sides, we’ll have the equation in terms of Heat Capacity.
$\dfrac{{dQ}}{{dT}} = \dfrac{{dU}}{{dT}} + \dfrac{{PdV}}{{dT}} \Rightarrow C = {C_V}$