Question

(3)

Calculate the entropy change when a sample of argon at 25 °C and 1.00 atm in a container
of volume 500 cm3 is compressed to 50 cm3 and cooled to -25 °C.
For argon, Cpm = 20.786) K-Imol-1. Assume that argon behaves perfectly.
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Answer 1

`\Delta s=-23.0(J)/(mol*K)`

Step-by-step explanation:

Hello!

In this case, since the entropy change when an ideal gas undergo both a volume and temperature change is computed by:

`\Delta s =Cp*ln((T_2)/(T_1) )-R*ln((P_2)/(P_1) )`

Whereas the molar Cp is 20.786 J/(mol*K) and the final pressure is computed via the Boyle's law:

`P_2=(1.00atm*500cm^3)/(50cm^3)=10 atm`

Thus, we plug in the data (temperatures in Kelvin) to obtain:

`\Delta s =20.786(J)/(mol*K) *ln[((-25+273)K)/((25+273)K) ]-8.3145(J)/(mol*K)*ln((10atm)/(1atm) )\\\\\Delta s=-23.0(J)/(mol*K)`

Best regards!