Heat capacity

Heat capacity, $C$ , is the ratio of heat transferred to a system from its surroundings to the temperature change of the system due to the transfer. In other words, the heat capacity of a system or substance is the amount of heat the system or substance can hold per Kelvin. The transfer of heat to a system can take place at either constant pressure or constant volume, resulting in two types of heat capacities, $C_p$ and $C_V$ respectively.

To define the heat capacity of a system at constant pressure, we begin with eq30 of the previous article or its differential form:

$dH=\delta q\;\;\;\;\;\;\;\;(34)$

where the symbol $\delta$ is to signify that $q$ is a path function.

Question

Does eq34 say that a state function is equal to a path function?

Answer

No, eq34 means that the value of $dH$ is equal to the change in $q$ along a path, but is always the same regardless of the path taken .

Dividing both sides of eq34 by $dT$ ,

$\frac{dH}{dT}=\frac{\delta q}{dT}\;\;\;\;\;\;\;\;(35)$

For a process at constant pressure, we rewrite eq35 as:

$\left(\frac{\partial H}{\partial T}\right)_p=\frac{\delta q_p}{dT}\;\;\;\;\;\;\;\;(36)$

The difference in symbols on the LHS versus the RHS of eq36 for indicating the process at constant pressure is due to $H$ being a state function and $q$ , a path function. According to the definition of heat capacity in the first paragraph of the article,

$\frac{\delta q_p}{dT}=\left(\frac{\partial H}{\partial T}\right)_p=C_p\;\;\;\;\;\;\;\;(37)$

Since the quantities $\delta q$ and $dT$ are easily found for different systems (or substances) via experiments like the neutralisation reaction mentioned in the previous article, the heat capacities for various systems or substances at constant pressure are easily determined. The molar heat capacity of a substance at constant pressure is:

$C_{p,m}=\left(\frac{\partial H_m}{\partial T}\right)_p\;\;\;\;\;\;\;\;(38)$

and the specific heat capacity of a substance at constant pressure, $C_{p,shc}$ , is found by dividing eq37 with the mass, $M$ , of the substance:

$dH=MC_{p,shc}dT\;\;\;\;at \;constant\; pressure\;\;\;\;(39)$

Eq39 is a useful equation to calculate the change in enthalpy of a system or substance. From eq37, the heat capacity of a substance at constant pressure is the gradient of a curve on a plot of enthalpy against temperature at constant pressure. If the curve is a straight line, $C_{p}$ is a constant at all temperatures. In reality, it is a function of temperature.

For experiments over a small temperature range, $C_{p}$ may be assumed to be constant to simplify calculations. For experiments that are carried out over a large temperature range, we need to find an expression for $C_p(T)$ . This is done by using a power series to generate curves that fit experimental values of $C_{p}$ for different substances on $C_{p}$ versus temperature plots:

$C_{p}(T)=a+bT+cT^2+dT^3+\cdots$

The set of constants $a$ , , $b$ , $c$ , $d$ etc. are specific to each substance and are the outcome of the polynomial regression. As the contribution of higher powers of $T$ to $C_{p}$ is small, the expression is fairly well represented by:

$C_{p}(T)=a+bT+cT^2+dT^3\;\;\;\;\;\;\;\;(40)$

Eq40 is then substituted in eq39 and the resultant equation is integrated throughout to obtain:

$\Delta H=M\left(a\Delta T+\frac{1}{2}b\Delta T^2+\frac{1}{3}c\Delta T^3+\frac{1}{4}d\Delta T^4\right)\;\;\;\;at\;constant\;pressure$

Using the same logic, to define the heat capacity of a system at constant volume, we begin with eq31 of the previous article or its differential form $dU=\delta q_{sys}$ , resulting in

$\frac{\delta q_V}{dT}=C_{V,m}=\left(\frac{\partial U_m}{\partial T}\right)_V\;\;\;\;\;\;\;\;(41)$

$\delta q=C_{V,m}dT\;\;\;or\;\;\;dU=MC_{V,shc}dT\;\;\;\;at \;constant\; volume\;\;\;\;(42)$

and

$\Delta U=M\left(a\Delta T+\frac{1}{2}b\Delta T^2+\frac{1}{3}c\Delta T^3+\frac{1}{4}d\Delta T^4\right)\;\;\;\;at\;constant\;volume\;\;\;\;(43)$

Since the value of the heat capacity of a system depends on whether we increase the amount of heat of the system at constant volume or at constant pressure, heat capacity $C$ is a path function. However, if the path is chosen, then $C_p$ or $C_V$ are state functions.

Finally, we shall show that $\left(\frac{\partial U}{\partial T}\right)_V=\frac{dU}{dT}=C_V(T)$ and $\left(\frac{\partial H}{\partial T}\right)_p=\frac{dH}{dT}=C_p(T)$ for a system containing an ideal gas.

The internal energy of a gas is the sum of the molecules’ translational energy $U_{tr}$ , rotational energy $U_{rot}$ , vibrational energy $U_{vib}$ , electronic transition energy $U_{el}$ , intermolecular forces of interaction $U_{im}$ and rest-mass energy of electrons and nuclei $U_{rest}$ :

$U=U_{tr}+U_{rot}+U_{vib}+U_{el}+U_{im}+U_{rest}\;\;\;\;\;\;(43a)$

$U_{rest}$ is constant for any gas, while $U_{el}$ is constant at $T<5000K$ and if no chemical reaction takes place. For an ideal gas, $U_{im}$ is zero, leaving the internal energy of an ideal gas as:

$U=U_{tr}+U_{rot}+U_{vib}+constant$

It can be shown that $U_{tr}$ , $U_{rot}$ and $U_{vib}$ are functions of temperature (independent of volume). Since the internal energy of an ideal gas is only dependent on temperature, $\left(\frac{\partial U}{\partial T}\right)_V=\frac{dU}{dT}$ . With $\left(\frac{\partial U}{\partial T}\right)_V=C_V$ , we write, for an ideal gas:

$\frac{dU}{dT}=C_V(T)\;\;\;\;or\;\;\;\;dU=C_V(T)dT\;\;\;\;\;(44)$

Eq44 is a useful relation that is applicable to all reversible processes that involves an ideal gas (not just constant volume processes because the internal energy of an ideal gas is only dependent on temperature). Irreversible processes have systems with poorly defined $T$ and therefore have internal energy changes that cannot be accurately predicted by eq44. For the enthalpy of an ideal gas,

$H=U+pV=U+nRT\;\;\;\;\;\;\;\;(45)$

As $U$ of an ideal gas is only dependent on temperature, the enthalpy of an ideal gas according to eq45 is also only dependent on temperature. Using the same logic as above, $\left(\frac{\partial H}{\partial T}\right)_p=\frac{dH}{dT}$ and

$\frac{dH}{dT}=C_p(T)\;\;\;\;or\;\;\;\;dH=C_p(T)dT\;\;\;\;\;(46)$

Question

Show that $C_p-C_V=nR$ for a perfect gas (one that obeys the ideal gas law and exhibits a heat capacity that is independent of temperature) using the above diagram where

AC: reversible isothermal process
AB: reversible isochoric process
BC: reversible isobaric processes

Answer

The change in internal energy $\Delta U$ from point A to C along the isotherm AC is zero for an ideal gas. Since $U$ is a state function, $\Delta U$ for the process AC and ABC is the same: