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One mole of a monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1L to 3L. The final temperature of the gas is:
- a)T
- b)T/3(5R/2)
- c)T – 2/(1.5 × 0.082)
- d)T + 2/(1.5 × 0.082)
Correct answer is option 'C'. Can you explain this answer?
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One mole of a monoatomic ideal gas expands adiabatically at initial te...
Since the process is adiabatic, we know that Q = 0. Therefore, we can use the first law of thermodynamics to find the final temperature:
ΔU = W
where ΔU is the change in internal energy and W is the work done on the gas. For an adiabatic process, ΔU = -W, so we have:
W = -PextΔV = -1 atm * (3L - 1L) = -2 atm L
ΔU = -W = 2 atm L
For a monoatomic ideal gas, the internal energy is given by:
U = (3/2) * nRT
where n is the number of moles, R is the gas constant, and T is the temperature. Therefore, we have:
ΔU = (3/2) * nRΔT
2 atm L = (3/2) * nRΔT
ΔT = (4/3) * (2 atm L) / (nR)
We can use the ideal gas law to find n:
PV = nRT
n = PV/RT = (1 atm * 1 L) / (R * T)
Substituting this expression for n into our equation for ΔT, we get:
ΔT = (4/3) * (2 atm L) / [(1 atm * 1 L) / T] = 8/3 * T
Therefore, the final temperature is:
Tf = Ti + ΔT = T + 8/3 * T = 11/3 * T
So the answer is (c) T.
ΔU = W
where ΔU is the change in internal energy and W is the work done on the gas. For an adiabatic process, ΔU = -W, so we have:
W = -PextΔV = -1 atm * (3L - 1L) = -2 atm L
ΔU = -W = 2 atm L
For a monoatomic ideal gas, the internal energy is given by:
U = (3/2) * nRT
where n is the number of moles, R is the gas constant, and T is the temperature. Therefore, we have:
ΔU = (3/2) * nRΔT
2 atm L = (3/2) * nRΔT
ΔT = (4/3) * (2 atm L) / (nR)
We can use the ideal gas law to find n:
PV = nRT
n = PV/RT = (1 atm * 1 L) / (R * T)
Substituting this expression for n into our equation for ΔT, we get:
ΔT = (4/3) * (2 atm L) / [(1 atm * 1 L) / T] = 8/3 * T
Therefore, the final temperature is:
Tf = Ti + ΔT = T + 8/3 * T = 11/3 * T
So the answer is (c) T.
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One mole of a monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1L to 3L. The final temperature of the gas is:a)Tb)T/3(5R/2)c)T – 2/(1.5 × 0.082)d)T + 2/(1.5 × 0.082)Correct answer is option 'C'. Can you explain this answer?
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One mole of a monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1L to 3L. The final temperature of the gas is:a)Tb)T/3(5R/2)c)T – 2/(1.5 × 0.082)d)T + 2/(1.5 × 0.082)Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about One mole of a monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1L to 3L. The final temperature of the gas is:a)Tb)T/3(5R/2)c)T – 2/(1.5 × 0.082)d)T + 2/(1.5 × 0.082)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1L to 3L. The final temperature of the gas is:a)Tb)T/3(5R/2)c)T – 2/(1.5 × 0.082)d)T + 2/(1.5 × 0.082)Correct answer is option 'C'. Can you explain this answer?.
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