calculate the work done when one gram molecule of gas expand isothermally at 27 degree Celsius to three times its initial volume ( R = 8.3 joule per degree mole)
Answers (2)
6th Nov, 2020
Hello there!
Here is the solution regarding your question.
Given,
T = 27°C [which is equal to 300 K].
Work done = 2.303 RT log Vf / Vi [ for isothermal process ].
W = 2.303 x 8.31 x 300 x log 3/1 [ log 3 = 0.48 ].
W = 19.1 x 300 x 0.48
W = 19.1 x 144
W = 2750.4 Joule
W = 2.75 x 10³ Joule.
Hence, the work done is 2.75 x 10³ Joule.
Hope you got your answer.
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Comments (0)
5th Nov, 2020
Dear student
Temp = 27 + 273 = 300k
V1= V
V2= 3V
W= 2.3038.3300log 10 (3)
= 2735 .5623 joule.
Temp = 27 + 273 = 300k
V1= V
V2= 3V
W= 2.3038.3300log 10 (3)
= 2735 .5623 joule.
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