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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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