Thermal Dynamics: Heat Capacity & Joule-Thomson

Heat Capacity

A system with mass $\mathit{m}$ will experience a temperature increase from ${\mathit{T}}_1$ to ${\mathit{T}}_2$ if $\mathit{q}$ units of heat are applied to it. Then 

$\mathit{q}\propto \left({\mathit{T}}_2-{\mathit{T}}_1\right)\mathit{m}$

or $\mathit{q}=\mathit{mC}\left({\mathit{T}}_2-{\mathit{T}}_1\right)$

where $\mathit{C}$ is known as the heat capacity of the system.

Heat capacity at constant volume

The heat needed to raise the system's temperature by one degree while keeping the volume constant is known as the heat capacity at constant volume, or ${\mathit{c}}_{\mathit{v}}$.

When only one mole of gas is considered, the quantity of heat needed to increase the temperature of one mole of gas by one degree at constant volume is known as the molar heat capacity.

Molar heat at constant volume is represented as ${\mathit{C}}_{\mathit{v}}$.

We know, from first law of thermodynamics

$\mathit{dq}=\mathit{dE}+\mathit{PdV}$

Dividing both sides by $\mathit{dT}$, we get

$\frac{\mathit{dq}}{\mathit{dT}}=\frac{\mathit{dE}}{\mathit{dT}}+\frac{\mathit{PdV}}{\mathit{dT}}$

At constant volume $\mathit{dV}=0$, 

$\left(\frac{\mathit{\delta q}}{\mathit{\delta T}}\right)\mathit{v}=\left(\frac{\partial \mathit{E}}{\partial \mathit{T}}\right)\mathit{v}={\mathit{C}}_{\mathit{v}}$

Heat capacity at constant pressure

Heat capacity under constant pressure is the amount of heat needed to raise the temperature by one degree while maintaining the same pressure, denoted by ${\mathit{c}}_{\mathit{p}}$.

The molar heat capacity at constant pressure, ${\mathit{C}}_{\mathit{p}}$, is the value for a system containing one mole of a substance.

Using the first law of thermodynamics, 

$\mathit{\Delta q}=\mathit{dE}+\mathit{PdV}$

dividing both sides by $\mathit{dT}$ we get

$\frac{\partial \mathit{q}}{\partial \mathit{T}}=\frac{\mathit{dE}}{\mathit{dT}}+\frac{\mathit{PdV}}{\mathit{dT}}$

We know $\mathit{H}=\mathit{E}+\mathit{PV}$

$\frac{\mathit{dH}}{\mathit{dT}}=\frac{\mathit{dE}}{\mathit{dT}}+\frac{\mathit{d}\left(\mathit{PV}\right)}{\mathit{dT}}$

Keeping the pressure constant

$\left(\frac{\partial \mathit{H}}{\partial \mathit{T}}\right)\mathit{p}=\left(\frac{\partial \mathit{E}}{\partial \mathit{T}}\right)\mathit{p}+\mathit{P}\left(\frac{\partial \mathit{V}}{\partial \mathit{T}}\right)\mathit{p}$

Hence, $\left(\frac{\partial \mathit{q}}{\partial \mathit{T}}\right)\mathit{p}=\left(\frac{\partial \mathit{H}}{\partial \mathit{T}}\right)\mathit{p}={\mathit{C}}_{\mathit{p}}$

From above, it is clear that two heat capacities are not equivalent and that ${\mathit{C}}_{\mathit{p}}$ exceeds ${\mathit{C}}_{\mathit{v}}$ by a quantity that is related to the amount of work done. At constant pressure, a portion of the heat received by the system is utilized to enhance its internal energy, and the other portion is used by the system to perform work. Since the system is not performing any work when operating at constant volume, all of the heat absorbed is used to boost its internal energy.

Thus, ${\mathit{C}}_{\mathit{p}}$ is greater than ${\mathit{C}}_{\mathit{v}}$.