KTGMedium
Question
When a system is taken from state ′a′ to state ′b′ along the path ′acb′, it is found that a quantity of heat Q = 200 J is absorbed by the system and a work W = 80J is done by it. Along the path ′adb′, Q = 144J. The work done along the path ′adb′ is
Options
A.6J
B.12J
C.18J
D.24J
Solution
ᐃU = same is both process
Qacb − Wacb = Qadb − Wadb .
200 − 80 = 144 − Wadb. Wadb = 24 J.
Qacb − Wacb = Qadb − Wadb .
200 − 80 = 144 − Wadb. Wadb = 24 J.
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