# 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?

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

#DeltaS ("const. P") -= int_(T_1)^(T_2) (C_P(T))/TdT#

#= int_(T_1)^(T_2) [20.78600/T + 2.825911 xx 10^(-10) - 1.464191 xx 10^(-13)T + 1.092131 xx 10^(-17) T^2 - (3.661371 xx 10^(-2))/T^3] dT#

#~~ int_(T_1)^(T_2) 20.78600/T dT# when ignoring the higher-order terms. The most exact answer is

#28.8_(155)color(white)(.)"J/K"# , and the approximated answer is#28.8_(35)# #"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:

#((delS)/(delT))_P# ,the partial derivative of

#S# (entropy) with respect to#T# (temperature) at constant pressure (#P# ).

Recall that the differential **entropy** is given by:

#dS = (deltaq_(rev))/T# #" "bb((1))# where

#"rev"# indicates the process is reversible, i.e. that is doneinfinitesimally slowly.

At **constant pressure**, the differential heat flow

#dH = overbrace(dE)^"Internal Energy" + d(PV)#

#= overbrace(deltaq_(rev) + deltaw_(rev))^("First Law of Thermodynamics") + PdV + VdP#

#= deltaq_(rev) cancel(- PdV + PdV) + VdP#

#= deltaq_(rev) + cancel(VdP)^(0)# at const. pressure,where

#w_(rev)# is thereversible workdone from the perspective of the system.

Furthermore, by definition, for a *reversible process*, the **constant-pressure heat capacity**

#C_P = ((delH)/(delT))_P = ((delq_(rev))/(delT))_P# #" "bb((2))#

Therefore, we use

#T((delS)/(delT))_P = ((delq_(rev))/(delT))_P = C_P# #" "bb((3))#

As a result, we have that:

#color(blue)(barul(|stackrel(" ")(" "((delS)/(delT))_P = C_P/T" ")|))# #" "# #ul"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_((1))^((2))dS = int_(T_1)^(T_2) C_P/TdT#

Assuming that the constant-pressure heat capacity *stays constant in the temperature range*, we can pull it out of the integral to get:

#color(blue)(DeltaS) ~~ C_Pln(T_2/T_1)#

#~~ nbarC_Pln(T_2/T_1)# #" "" "# (where#barC_P = C_P/n# )

#~~# #cancel"1 mol" cdot "20.8 J/"cancel"mol"cdot"K" cdot ln("1200 K"/"300 K")#

#~~# #color(blue)(ul"28.8 J/K")#