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

Answer:

$\Delta s=-23.0\frac{J}{mol*K}$

Explanation:

Hello!

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

$\Delta s =Cp*ln(\frac{T_2}{T_1} )-R*ln(\frac{P_2}{P_1} )$

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

$P_2=\frac{1.00atm*500cm^3}{50cm^3}=10 atm$

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

$\Delta s =20.786\frac{J}{mol*K} *ln[\frac{(-25+273)K}{(25+273)K} ]-8.3145\frac{J}{mol*K}*ln(\frac{10atm}{1atm} )\\\\\Delta s=-23.0\frac{J}{mol*K}$

Best regards!