# What is the entropy change at #"273 K"# when a perfect gas is compressed at #"1 bar"# to a volume of #"0.011 m"^3# at #"10 bar"#?

##### 1 Answer

I got

Consider entropy as *a function of temperature and pressure*. We can write the **total differential** as:

#dS(T,P) = ((delS)/(delT))_PdT + ((delS)/(delP))_TdP#

- The first derivative term,
#((delS)/(delT))_P# , describes the change in entropy due to the change in*temperature*at a constant*pressure*. - The second derivative term,
#((delS)/(delP))_T# , describes the change in entropy due to the change in*pressure*at a constant*temperature*.

Notice that the problem asks for a change in pressure but only tells you one temperature. That is, the temperature is ** constant**. So, we're searching for how to evaluate the second derivative seen above.

Recall that the **Gibbs' free energy** is commonly regarded as a function of the temperature and pressure, i.e. *Maxwell Relation*:

#dG = -SdT + VdP#

Since

#((delS)/(delP))_T = -((delV)/(delT))_P#

Now that we've established the relationship, we have an expression to use in an equation we know: the **ideal gas law**.

#PV = nRT#

The partial derivative of the volume with respect to the temperature at *constant* pressure is then:

#-(del)/(delT)[V]_P = -(del)/(delT)[(nRT)/P]_P#

#= -(nR)/P(del)/(delT)[T]_P#

#= -(nR)/P#

The next thing to do is *integrate* over the two pressures.

#DeltaS = int_(P_1)^(P_2)((delS)/(delP))_TdP#

#= -int_(P_1)^(P_2)((delV)/(delT))_PdP#

#= -int_(P_1)^(P_2)(nR)/PdP#

#= -nRint_(P_1)^(P_2)1/PdP#

The integral of

#DeltaS = -nR(ln|P_2| - ln|P_1|)#

#= -nRln|P_2/P_1|#

Since you are only given the molar volume and not the mols, we have to assume that at **the starting molar volume has to be calculated** at

#V/n = (RT)/P#

#= (("0.083145 L"cdot"atm/mol"cdot"K")("273 K"))/("1 bar")#

#=# #"22.700 L/mol"#

Therefore, the *mols of perfect gas* present are:

#n = "1 mol"/(22.700 cancel"L") xx 0.011 cancel("m"^3) xx ((cancel"1 dm")/(0.1 cancel"m"))^3 xx cancel("1 L")/cancel("1 dm"^3)#

#=# #"0.485 mols gas"#

Finally, we can calculate the **change in entropy** at *constant* temperature due to a *change* in pressure:

#color(blue)(DeltaS) = -(0.485 cancel"mols")("8.314472 J/"cancel"mol"cdot"K")ln("10 bar"/"1 bar")#

#= color(blue)(-"9.28 J/K")#