Two moles of an ideal monoatomic gases are allowed to expand adiabatically and reversibily from 300k and 200k. The work done in the system is
Answers
Answered by
42
W= nCv[T2-T1]
W= 2 * 12.5×[200-300]
W= -2500
W= -2.5 KJ
Answered by
26
Hello dear,
● Answer -
W = -2.5 kJ
● Explaination-
# Given-
n = 2
T1 = 300 K
T2 = 200 K
Cv = 12.5 J/molK
# Solution-
Work done in adiabatic process is calculated by-
W = nCv∆T
W = 2 × 12.5 × (200-300)
W = 25 × (-100)
W = -2500 J
W = -2.5 kJ
Therefore, work done in the system is -2.5 kJ.
Hope this helps...
● Answer -
W = -2.5 kJ
● Explaination-
# Given-
n = 2
T1 = 300 K
T2 = 200 K
Cv = 12.5 J/molK
# Solution-
Work done in adiabatic process is calculated by-
W = nCv∆T
W = 2 × 12.5 × (200-300)
W = 25 × (-100)
W = -2500 J
W = -2.5 kJ
Therefore, work done in the system is -2.5 kJ.
Hope this helps...
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