Sulphur dioxide is circulated as refrigerant in a small refrigerator. SO2 gas at a pressure of 5 bar and temperature 340 K is to be cooled at a constant volume of 0.142 m3, to 293 K as part of the refrigeration cycle. Calculate (a) The heat liberated, (b) The work done by the gas on cooling, (c) The final pressure attained on cooling and (d) The change in enthalpy. Sulphur dioxide may be treated as an ideal gas. The specific heat (J/mol K) is found to vary with temperature (K) according to Cp = 25.736 + 5.796 X 10-2 T – 3.8112 x 10-5 T2 + 8.612 x 10-9 T3

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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(a) 38.301 x 10^3J (b) 0 (c) 4.31 bar (d) – 47.817 x 10^3 J

Sulphur dioxide is circulated as refrigerant in a small refrigerator. SO2 gas at a
pressure of 5 bar and temperature 340 K is to be cooled at a constant volume of
0.142 m3, to 293 K as part of the refrigeration cycle. Calculate
(a) The heat liberated,
(b) The work done by the gas on cooling,
(c) The final pressure attained on cooling and
(d) The change in enthalpy.
Sulphur dioxide may be treated as an ideal gas. The specific heat (J/mol K) is
found to vary with temperature (K) according to
CP = 25.736 + 5.796 X 10-2 T – 3.8112 x 10-5 T2 + 8.612 x 10-9 T3
Transcribed Image Text:Sulphur dioxide is circulated as refrigerant in a small refrigerator. SO2 gas at a pressure of 5 bar and temperature 340 K is to be cooled at a constant volume of 0.142 m3, to 293 K as part of the refrigeration cycle. Calculate (a) The heat liberated, (b) The work done by the gas on cooling, (c) The final pressure attained on cooling and (d) The change in enthalpy. Sulphur dioxide may be treated as an ideal gas. The specific heat (J/mol K) is found to vary with temperature (K) according to CP = 25.736 + 5.796 X 10-2 T – 3.8112 x 10-5 T2 + 8.612 x 10-9 T3
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why does the change in volume in enthalpy formula is not 0, considering the change in volume in work is 0

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