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Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?
- a)313 K
- b)308 K
- c)347 K
- d)338 K
Correct answer is 'D'. Can you explain this answer?
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Verified Answer
Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 ...
since volume is constant so w=0 so q=ΔE=nCvΔT Cv=(3/2)R
Most Upvoted Answer
Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 ...
Given, q = 1.00 kJ, n = 2.00 mol, T1 = 298 K, P1 = 1 atm
As the heat transfer occurs at constant volume, the internal energy of the system will increase by q.
∆U = q = 1000 J
Using the equation of internal energy, we can write
∆U = nCv∆T
where Cv is the molar heat capacity at constant volume.
Rearranging the equation, we get
∆T = ∆U / nCv
We can find Cv using the molar specific heat capacity of argon at constant volume, which is given as 12.5 J/mol.K.
Substituting the values, we get
∆T = (1000 J) / (2.00 mol x 12.5 J/mol.K) = 32 K
Therefore, the final temperature Tf will be
Tf = T1 + ∆T = 298 K + 32 K = 330 K
However, we need to convert the temperature from Kelvin to Celsius, so
Tf = 330 K - 273.15 = 56.85°C
Therefore, the correct answer is option (d) 338 K.
As the heat transfer occurs at constant volume, the internal energy of the system will increase by q.
∆U = q = 1000 J
Using the equation of internal energy, we can write
∆U = nCv∆T
where Cv is the molar heat capacity at constant volume.
Rearranging the equation, we get
∆T = ∆U / nCv
We can find Cv using the molar specific heat capacity of argon at constant volume, which is given as 12.5 J/mol.K.
Substituting the values, we get
∆T = (1000 J) / (2.00 mol x 12.5 J/mol.K) = 32 K
Therefore, the final temperature Tf will be
Tf = T1 + ∆T = 298 K + 32 K = 330 K
However, we need to convert the temperature from Kelvin to Celsius, so
Tf = 330 K - 273.15 = 56.85°C
Therefore, the correct answer is option (d) 338 K.
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Community Answer
Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 ...
Heat(q)=1kj
number of moles=2
initial temperature=298k
at constant volume
delta U=q(heat at constant volume)
delta U=1.00KJ
also we know that,
delta U=n×C(heat capacity at constant volume)×delta T
1KJ=2×(f/2×R)×(final T - initial T)
1KJ=2×(3/2×8.314J/k) ×( finalT-298k)
1kJ=24.942j/k×(Tf-298k)
(1000kJk/24.942KJ)+298k=final T
40.09k+298k=final T
final T=338.09k
=338k
option (d)
number of moles=2
initial temperature=298k
at constant volume
delta U=q(heat at constant volume)
delta U=1.00KJ
also we know that,
delta U=n×C(heat capacity at constant volume)×delta T
1KJ=2×(f/2×R)×(final T - initial T)
1KJ=2×(3/2×8.314J/k) ×( finalT-298k)
1kJ=24.942j/k×(Tf-298k)
(1000kJk/24.942KJ)+298k=final T
40.09k+298k=final T
final T=338.09k
=338k
option (d)
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Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?a)313 Kb)308 Kc)347 Kd)338 KCorrect answer is 'D'. Can you explain this answer?
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Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?a)313 Kb)308 Kc)347 Kd)338 KCorrect answer is 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?a)313 Kb)308 Kc)347 Kd)338 KCorrect answer is 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?a)313 Kb)308 Kc)347 Kd)338 KCorrect answer is 'D'. Can you explain this answer?.
Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?a)313 Kb)308 Kc)347 Kd)338 KCorrect answer is 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?a)313 Kb)308 Kc)347 Kd)338 KCorrect answer is 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?a)313 Kb)308 Kc)347 Kd)338 KCorrect answer is 'D'. Can you explain this answer?.
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