Starting with the differential form of the Helmholtz free energy F, derive and thus OF Ꭲ OF = -S and = -P Ꮩ V (as = av Use the expression derived in part a to determine the change in entropy when one mol of mercury is compressed from an initial start = 1 x 104 m³ to a final volume Vend = 2 × 10 5 m³. Assume the mercury obeys the Van der Waals equation for a non-ideal gas, RT a P = V-b for mercury: a = 0.8200m Pa and b = 1.696 x 10 m³.
Starting with the differential form of the Helmholtz free energy F, derive and thus OF Ꭲ OF = -S and = -P Ꮩ V (as = av Use the expression derived in part a to determine the change in entropy when one mol of mercury is compressed from an initial start = 1 x 104 m³ to a final volume Vend = 2 × 10 5 m³. Assume the mercury obeys the Van der Waals equation for a non-ideal gas, RT a P = V-b for mercury: a = 0.8200m Pa and b = 1.696 x 10 m³.
Chapter4: The Second Law Of Thermodynamics
Section: Chapter Questions
Problem 86CP: A cylinder contains 500 g of helium at 120 atm and 20 . The valve is leaky, and all the gas slowly...
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
Transcribed Image Text:Starting with the differential form of the Helmholtz free energy F, derive
and thus
OF
Ꭲ
OF
= -S
and
= -P
Ꮩ
V
(as
=
av
Use the expression derived in part a to determine the change in entropy when one mol of
mercury is compressed from an initial start = 1 x 104 m³ to a final volume Vend = 2 × 10 5 m³.
Assume the mercury obeys the Van der Waals equation for a non-ideal gas,
RT
a
P =
V-b
for mercury: a = 0.8200m Pa and b = 1.696 x 10 m³.
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