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1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to ? (Given: CV = 20 J mol–1K–1):
Correct answer is '150'. Can you explain this answer?
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1 mole of an ideal gas is allowed to expand reversibly and adiabatical...
From first law of thermodynamics,
ΔU=q+W
As we know that, for adiabatic condition,
q=0
∴ΔU=W.....(1)
At constant volume,
ΔU=nCvΔT.....(2)
From eqn(1)&(2), we have
W=nCvΔT.....(3)
Given:-
Ti=27℃=(27+273)K=300K
Tf=T(say)=?
ΔT=Tf−Ti=(T−300)
Cv=20J/K −mol
W=−3kJ=−3×103J[∵Work done by the gas is negative]
Substituting all these values in eqn(3), we have
−3000=1×20×(T−300)
T−300=−150
⇒T=300−150=150K
Hence the final temperature is 150K.
Most Upvoted Answer
1 mole of an ideal gas is allowed to expand reversibly and adiabatical...
°C (300 K) to a temperature of 127°C (400 K). Calculate the change in entropy of the gas during this process.
We can use the formula for the change in entropy of an ideal gas undergoing an adiabatic expansion:
ΔS = nCv ln(T2/T1)
where ΔS is the change in entropy, n is the number of moles of gas (in this case, 1), Cv is the molar specific heat at constant volume, and T1 and T2 are the initial and final temperatures, respectively.
We can find Cv using the ideal gas law and the definition of Cv:
Cv = (dU/dT)v
where U is the internal energy of the gas and v is the volume. For an ideal gas, U depends only on temperature, so we can write:
dU = Cv dT
Substituting this into the ideal gas law, we get:
PV = nRT = (nCv)T
Solving for Cv, we get:
Cv = (nR)/(γ-1)
where γ is the ratio of specific heats (Cp/Cv) for the gas. For a monatomic gas, γ = 5/3, so:
Cv = (3/2)R
Substituting all of these values into the formula for ΔS, we get:
ΔS = (1)(3/2R) ln(400/300) = (3/2)R ln(4/3)
Using R = 8.31 J/mol*K, we get:
ΔS = (3/2)(8.31) ln(4/3) ≈ 5.7 J/K
Therefore, the change in entropy of the gas during this process is approximately 5.7 J/K.
We can use the formula for the change in entropy of an ideal gas undergoing an adiabatic expansion:
ΔS = nCv ln(T2/T1)
where ΔS is the change in entropy, n is the number of moles of gas (in this case, 1), Cv is the molar specific heat at constant volume, and T1 and T2 are the initial and final temperatures, respectively.
We can find Cv using the ideal gas law and the definition of Cv:
Cv = (dU/dT)v
where U is the internal energy of the gas and v is the volume. For an ideal gas, U depends only on temperature, so we can write:
dU = Cv dT
Substituting this into the ideal gas law, we get:
PV = nRT = (nCv)T
Solving for Cv, we get:
Cv = (nR)/(γ-1)
where γ is the ratio of specific heats (Cp/Cv) for the gas. For a monatomic gas, γ = 5/3, so:
Cv = (3/2)R
Substituting all of these values into the formula for ΔS, we get:
ΔS = (1)(3/2R) ln(400/300) = (3/2)R ln(4/3)
Using R = 8.31 J/mol*K, we get:
ΔS = (3/2)(8.31) ln(4/3) ≈ 5.7 J/K
Therefore, the change in entropy of the gas during this process is approximately 5.7 J/K.
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1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to ? (Given: CV = 20 J mol–1K–1):Correct answer is '150'. Can you explain this answer?
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1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to ? (Given: CV = 20 J mol–1K–1):Correct answer is '150'. 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 1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to ? (Given: CV = 20 J mol–1K–1):Correct answer is '150'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to ? (Given: CV = 20 J mol–1K–1):Correct answer is '150'. Can you explain this answer?.
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