# Calculate the change in entropy when "1 mol" of argon gas is heated from "300 K" to "1200 K", assuming that C_P = "20.8 J/mol"cdot"K" and is constant in the temperature range?

##### 1 Answer
Aug 7, 2017

About $\underline{{28.8}_{35} \textcolor{w h i t e}{.} \text{J/K}}$, assuming ${C}_{P}$ stays constant throughout the rather large temperature range. This is usually a good approximation for a gas in general, particularly at high temperatures.

If you really want to be exact, look up the Shomate equation for argon, and integrate that instead:

$\Delta S \left(\text{const. P}\right) \equiv {\int}_{{T}_{1}}^{{T}_{2}} \frac{{C}_{P} \left(T\right)}{T} \mathrm{dT}$

$= {\int}_{{T}_{1}}^{{T}_{2}} \left[\frac{20.78600}{T} + 2.825911 \times {10}^{- 10} - 1.464191 \times {10}^{- 13} T + 1.092131 \times {10}^{- 17} {T}^{2} - \frac{3.661371 \times {10}^{- 2}}{T} ^ 3\right] \mathrm{dT}$

$\approx {\int}_{{T}_{1}}^{{T}_{2}} \frac{20.78600}{T} \mathrm{dT}$

when ignoring the higher-order terms. The most exact answer is ${28.8}_{155} \textcolor{w h i t e}{.} \text{J/K}$, and the approximated answer is ${28.8}_{35}$ $\text{J/K}$. Pretty good!

DERIVING THE RELATIONSHIP BETWEEN ENTROPY AND HEAT CAPACITY

Well, the influence on the entropy due to a change in temperature at constant pressure is given by:

${\left(\frac{\partial S}{\partial T}\right)}_{P}$,

the partial derivative of $S$ (entropy) with respect to $T$ (temperature) at constant pressure ($P$).

Recall that the differential entropy is given by:

$\mathrm{dS} = \frac{\delta {q}_{r e v}}{T}$ $\text{ } \boldsymbol{\left(1\right)}$

where $\text{rev}$ indicates the process is reversible, i.e. that is done infinitesimally slowly.

At constant pressure, the differential heat flow $\delta q$ is equal to differential change in enthalpy, $\mathrm{dH}$, because from the first law of thermodynamics...

$\mathrm{dH} = {\overbrace{\mathrm{dE}}}^{\text{Internal Energy}} + d \left(P V\right)$

$= {\overbrace{\delta {q}_{r e v} + \delta {w}_{r e v}}}^{\text{First Law of Thermodynamics}} + P \mathrm{dV} + V \mathrm{dP}$

$= \delta {q}_{r e v} \cancel{- P \mathrm{dV} + P \mathrm{dV}} + V \mathrm{dP}$

$= \delta {q}_{r e v} + {\cancel{V \mathrm{dP}}}^{0}$ at const. pressure,

where ${w}_{r e v}$ is the reversible work done from the perspective of the system.

Furthermore, by definition, for a reversible process, the constant-pressure heat capacity ${C}_{P}$ is given by:

${C}_{P} = {\left(\frac{\partial H}{\partial T}\right)}_{P} = {\left(\frac{\partial {q}_{r e v}}{\partial T}\right)}_{P}$ $\text{ } \boldsymbol{\left(2\right)}$

Therefore, we use $\left(2\right)$ to modify $\left(1\right)$ to get:

$T {\left(\frac{\partial S}{\partial T}\right)}_{P} = {\left(\frac{\partial {q}_{r e v}}{\partial T}\right)}_{P} = {C}_{P}$ $\text{ } \boldsymbol{\left(3\right)}$

As a result, we have that:

$\textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "((delS)/(delT))_P = C_P/T" }}{|}}}}$$\text{ }$ $\underline{\text{at constant pressure}}$.

(If you have done the derivation before, you can feel free to just memorize this result.)

FINDING THE CHANGE IN ENTROPY

Keeping that in mind, separate the variables and integrate as follows:

${\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dS} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} / T \mathrm{dT}$

Assuming that the constant-pressure heat capacity ${C}_{P} \left(T\right)$ (normally a function of temperature) stays constant in the temperature range, we can pull it out of the integral to get:

$\textcolor{b l u e}{\Delta S} \approx {C}_{P} \ln \left({T}_{2} / {T}_{1}\right)$

$\approx n {\overline{C}}_{P} \ln \left({T}_{2} / {T}_{1}\right)$ $\text{ "" }$(where ${\overline{C}}_{P} = {C}_{P} / n$)

$\approx$ cancel"1 mol" cdot "20.8 J/"cancel"mol"cdot"K" cdot ln("1200 K"/"300 K")

$\approx$ $\textcolor{b l u e}{\underline{\text{28.8 J/K}}}$