as at 500 kPa and 80°C while the other side contains 1 m³ of Helium gas at 500 Pa and 25°C. Now thermal equilibrium is established in the cylinder as a result f heat transfer through the piston. Using constant specific heats at room emperature, determine the final equilibrium temperature in the cylinder. ties The gas constants and the constant volume specific heats are R= 0.2968 kPa.m³kg.K is c,= J/kg•°C_for N3, and R = 2.0769 kPa.m³/kg.K is c,= 3.1156 kJ/kg-°C for He
as at 500 kPa and 80°C while the other side contains 1 m³ of Helium gas at 500 Pa and 25°C. Now thermal equilibrium is established in the cylinder as a result f heat transfer through the piston. Using constant specific heats at room emperature, determine the final equilibrium temperature in the cylinder. ties The gas constants and the constant volume specific heats are R= 0.2968 kPa.m³kg.K is c,= J/kg•°C_for N3, and R = 2.0769 kPa.m³/kg.K is c,= 3.1156 kJ/kg-°C for He
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:10.Consider a well-insulated horizontal rigid cylinder that is divided into two
compartments by a piston that is free to move but does not allow either gas to
leak into the other side. Initially, one side of the piston contains 1 m3 of nitrogen
gas at 500 kPa and 80°C while the other side contains 1 m³ of Helium gas at 500
kPa and 25°C. Now thermal equilibrium is established in the cylinder as a result
of heat transfer through the piston. Using constant specific heats at room
temperature, determine the final equilibrium temperature in the cylinder.
Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m³/kg.K is c,=
0.743 kJ/kg-°C for N3, and R = 2.0769 kPa.m/kg.K is c,= 3.1156 kJ/kg-°C for He
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