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The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole?
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The Value of∆U for the transformation of one mole of an ideal gas from...
Calculation of ∆U for the Transformation of One Mole of an Ideal Gas
Given parameters:
Initial Temperature (T1) = 27°C = 300 K
Final Temperature (T2) = 327°C = 600 K
Initial Pressure (P1) = 1 atm
Final Pressure (P2) = 17 atm
CP = 20.9 + 0.042T J/K-mole
Formula to calculate ∆U:
∆U = nCP(T2 - T1)
Calculation of n (number of moles):
In the given problem, n = 1 mole
Calculation of CP:
CP = 20.9 + 0.042T J/K-mole
CP = 20.9 + 0.042(300) J/K-mole
CP = 20.9 + 12.6 J/K-mole
CP = 33.5 J/K-mole
Calculation of ∆U:
∆U = nCP(T2 - T1)
∆U = (1 mole)(33.5 J/K-mole)(600 K - 300 K)
∆U = 10,050 J/mol
Therefore, the value of ∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 327 degree Celsius and 17 atm, given that CP is equals to 20.9 + 0.042 T J/K-mole, is 10,050 J/mol.
Given parameters:
Initial Temperature (T1) = 27°C = 300 K
Final Temperature (T2) = 327°C = 600 K
Initial Pressure (P1) = 1 atm
Final Pressure (P2) = 17 atm
CP = 20.9 + 0.042T J/K-mole
Formula to calculate ∆U:
∆U = nCP(T2 - T1)
Calculation of n (number of moles):
In the given problem, n = 1 mole
Calculation of CP:
CP = 20.9 + 0.042T J/K-mole
CP = 20.9 + 0.042(300) J/K-mole
CP = 20.9 + 12.6 J/K-mole
CP = 33.5 J/K-mole
Calculation of ∆U:
∆U = nCP(T2 - T1)
∆U = (1 mole)(33.5 J/K-mole)(600 K - 300 K)
∆U = 10,050 J/mol
Therefore, the value of ∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 327 degree Celsius and 17 atm, given that CP is equals to 20.9 + 0.042 T J/K-mole, is 10,050 J/mol.
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The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole?
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The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole?.
The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Value of∆U for the transformation of one mole of an ideal gas from 27 degree C and 1 atm to 3 27 degree Celsius and 17 atm is given that CP is equals to( 20.9 0.042 T J/K -mole?.
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