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What is the final temperature (in kelvin) of 0.10 mole monoatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227°C ? (use R = 2 cal/K-mol)
Correct answer is '250'. Can you explain this answer?
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To solve this problem, we need to use the first law of thermodynamics for an adiabatic process:
ΔU = Q - W
Where:
ΔU is the change in internal energy
Q is the heat transferred to the system
W is the work done by the system
In an adiabatic process, Q = 0 because there is no heat transfer. Therefore, the equation simplifies to:
ΔU = -W
The change in internal energy (ΔU) can be calculated using the equation:
ΔU = (3/2)nRΔT
Where:
n is the number of moles
R is the ideal gas constant (8.314 J/(mol·K))
ΔT is the change in temperature
We can rearrange the equation to solve for ΔT:
ΔT = ΔU / [(3/2)nR]
Substituting the given values:
n = 0.10 mole
R = 8.314 J/(mol·K)
ΔU = -75 cal = -75 cal * 4.184 J/cal = -313.8 J
ΔT = -313.8 J / [(3/2)(0.10 mole)(8.314 J/(mol·K))]
≈ -313.8 J / (0.15 mole·K)
Now, we can calculate the final temperature (Tf) using the equation:
Tf = Ti + ΔT
Substituting the given initial temperature (Ti = 227 K):
Tf = 227 K + ΔT
Calculating ΔT using a negative value:
ΔT = -313.8 J / (0.15 mole·K)
≈ -2092 J / mole·K
Tf = 227 K + (-2092 J / mole·K)
≈ 227 K - 2092 K
≈ -1865 K
However, a negative temperature does not make physical sense in this context. Therefore, the final temperature cannot be determined with the given information.
ΔU = Q - W
Where:
ΔU is the change in internal energy
Q is the heat transferred to the system
W is the work done by the system
In an adiabatic process, Q = 0 because there is no heat transfer. Therefore, the equation simplifies to:
ΔU = -W
The change in internal energy (ΔU) can be calculated using the equation:
ΔU = (3/2)nRΔT
Where:
n is the number of moles
R is the ideal gas constant (8.314 J/(mol·K))
ΔT is the change in temperature
We can rearrange the equation to solve for ΔT:
ΔT = ΔU / [(3/2)nR]
Substituting the given values:
n = 0.10 mole
R = 8.314 J/(mol·K)
ΔU = -75 cal = -75 cal * 4.184 J/cal = -313.8 J
ΔT = -313.8 J / [(3/2)(0.10 mole)(8.314 J/(mol·K))]
≈ -313.8 J / (0.15 mole·K)
Now, we can calculate the final temperature (Tf) using the equation:
Tf = Ti + ΔT
Substituting the given initial temperature (Ti = 227 K):
Tf = 227 K + ΔT
Calculating ΔT using a negative value:
ΔT = -313.8 J / (0.15 mole·K)
≈ -2092 J / mole·K
Tf = 227 K + (-2092 J / mole·K)
≈ 227 K - 2092 K
≈ -1865 K
However, a negative temperature does not make physical sense in this context. Therefore, the final temperature cannot be determined with the given information.
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What is the final temperature (in kelvin) of 0.10 mole monoatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227°C ? (use R = 2 cal/K-mol)Correct answer is '250'. Can you explain this answer?
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What is the final temperature (in kelvin) of 0.10 mole monoatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227°C ? (use R = 2 cal/K-mol)Correct answer is '250'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about What is the final temperature (in kelvin) of 0.10 mole monoatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227°C ? (use R = 2 cal/K-mol)Correct answer is '250'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the final temperature (in kelvin) of 0.10 mole monoatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227°C ? (use R = 2 cal/K-mol)Correct answer is '250'. Can you explain this answer?.
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