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One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.
Correct answer is between '286,288'. Can you explain this answer?
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One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at...
Given that m = 1kg, R = 400 kJ/kgK, CV = 1000 J/kgK 11
P1 = 1 bar, T1= 300 K Since the gas is contained in a sealed rigid cylinder, and given that adiabatic process is done to the system, means no heat is transferred from/to the system, Q = 0 And we know from first law of thermo dynamics,
P1 = 1 bar, T1= 300 K Since the gas is contained in a sealed rigid cylinder, and given that adiabatic process is done to the system, means no heat is transferred from/to the system, Q = 0 And we know from first law of thermo dynamics,
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One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at...
Increase in entropy of the system during an adiabatic process can be calculated using the first law of thermodynamics and the equation for entropy change. The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added, and W is the work done.
In this case, the system is adiabatic, meaning there is no heat transfer (Q = 0). The work done on the system is given as 100 kJ.
1. Calculate the change in internal energy (ΔU):
ΔU = Q - W
ΔU = 0 - 100 kJ
ΔU = -100 kJ
2. Calculate the change in temperature (ΔT):
For an ideal gas, the change in internal energy is given by the equation:
ΔU = m * cv * ΔT
where m is the mass of the gas, cv is the specific heat at constant volume, and ΔT is the change in temperature.
Since the system is adiabatic, there is no heat transfer and ΔQ = 0. Therefore, the change in temperature can be calculated as:
ΔT = ΔU / (m * cv)
ΔT = -100 kJ / (1 kg * 1000 J/kg.K)
ΔT = -0.1 K
The change in temperature is negative, indicating a decrease in temperature.
3. Calculate the final temperature (T2):
T2 = T1 + ΔT
T2 = 300 K - 0.1 K
T2 = 299.9 K
4. Calculate the change in entropy (ΔS):
The change in entropy is given by the equation:
ΔS = cv * ln(T2 / T1)
where ln is the natural logarithm.
ΔS = 1000 J/kg.K * ln(299.9 K / 300 K)
ΔS ≈ 1000 J/kg.K * ln(0.9997)
ΔS ≈ 1000 J/kg.K * (-0.0003)
ΔS ≈ -0.3 J/K
The change in entropy is negative, indicating a decrease in entropy.
5. Calculate the entropy increase (|ΔS|):
The absolute value of the change in entropy is taken to consider the magnitude:
|ΔS| = |-0.3 J/K|
|ΔS| = 0.3 J/K
The increase in entropy of the system is approximately 0.3 J/K.
Hence, the correct answer is between 286 and 288 J/K.
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added, and W is the work done.
In this case, the system is adiabatic, meaning there is no heat transfer (Q = 0). The work done on the system is given as 100 kJ.
1. Calculate the change in internal energy (ΔU):
ΔU = Q - W
ΔU = 0 - 100 kJ
ΔU = -100 kJ
2. Calculate the change in temperature (ΔT):
For an ideal gas, the change in internal energy is given by the equation:
ΔU = m * cv * ΔT
where m is the mass of the gas, cv is the specific heat at constant volume, and ΔT is the change in temperature.
Since the system is adiabatic, there is no heat transfer and ΔQ = 0. Therefore, the change in temperature can be calculated as:
ΔT = ΔU / (m * cv)
ΔT = -100 kJ / (1 kg * 1000 J/kg.K)
ΔT = -0.1 K
The change in temperature is negative, indicating a decrease in temperature.
3. Calculate the final temperature (T2):
T2 = T1 + ΔT
T2 = 300 K - 0.1 K
T2 = 299.9 K
4. Calculate the change in entropy (ΔS):
The change in entropy is given by the equation:
ΔS = cv * ln(T2 / T1)
where ln is the natural logarithm.
ΔS = 1000 J/kg.K * ln(299.9 K / 300 K)
ΔS ≈ 1000 J/kg.K * ln(0.9997)
ΔS ≈ 1000 J/kg.K * (-0.0003)
ΔS ≈ -0.3 J/K
The change in entropy is negative, indicating a decrease in entropy.
5. Calculate the entropy increase (|ΔS|):
The absolute value of the change in entropy is taken to consider the magnitude:
|ΔS| = |-0.3 J/K|
|ΔS| = 0.3 J/K
The increase in entropy of the system is approximately 0.3 J/K.
Hence, the correct answer is between 286 and 288 J/K.
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One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.Correct answer is between '286,288'. Can you explain this answer?
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One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.Correct answer is between '286,288'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.Correct answer is between '286,288'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.Correct answer is between '286,288'. Can you explain this answer?.
One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.Correct answer is between '286,288'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.Correct answer is between '286,288'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, cn = 1000J/kg.K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.Correct answer is between '286,288'. Can you explain this answer?.
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