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One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).
Correct answer is '114.8'. Can you explain this answer?
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Verified Answer
One kg of air, initially at a temperature of 127°C, expands revers...
m = 1 kg; T1 = 127°C = 400 K
P = C; V2 = 2V1; R = 0.287 kJ/kgK
W = P1(V2 – V1)
= P1(2V1 – V1)
= P1 ·V1 = mRT1
= 114.8 kJ
Most Upvoted Answer
One kg of air, initially at a temperature of 127°C, expands revers...
°C, is heated at constant pressure until its temperature reaches 227°C. We are asked to find the change in heat transfer during this process.
To solve this problem, we can use the equation:
Q = mcΔT
Where:
Q = heat transfer
m = mass of the substance (in this case, air) = 1 kg
c = specific heat capacity of air at constant pressure = 1.005 kJ/kg°C (given)
ΔT = change in temperature = final temperature - initial temperature
First, we need to convert the temperatures from Celsius to Kelvin:
Initial temperature (Ti) = 127°C + 273.15 = 400.15 K
Final temperature (Tf) = 227°C + 273.15 = 500.15 K
Now, we can calculate the change in temperature:
ΔT = Tf - Ti
ΔT = 500.15 K - 400.15 K
ΔT = 100 K
Substituting the values into the equation, we get:
Q = (1 kg)(1.005 kJ/kg°C)(100 K)
Q = 100.5 kJ
Therefore, the change in heat transfer during this process is 100.5 kJ.
To solve this problem, we can use the equation:
Q = mcΔT
Where:
Q = heat transfer
m = mass of the substance (in this case, air) = 1 kg
c = specific heat capacity of air at constant pressure = 1.005 kJ/kg°C (given)
ΔT = change in temperature = final temperature - initial temperature
First, we need to convert the temperatures from Celsius to Kelvin:
Initial temperature (Ti) = 127°C + 273.15 = 400.15 K
Final temperature (Tf) = 227°C + 273.15 = 500.15 K
Now, we can calculate the change in temperature:
ΔT = Tf - Ti
ΔT = 500.15 K - 400.15 K
ΔT = 100 K
Substituting the values into the equation, we get:
Q = (1 kg)(1.005 kJ/kg°C)(100 K)
Q = 100.5 kJ
Therefore, the change in heat transfer during this process is 100.5 kJ.
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One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer?
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One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer?.
One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer?.
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One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer?, a detailed solution for One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer? has been provided alongside types of One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer? theory, EduRev gives you an
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