ThermodynamicsHard
Question
An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is
Options
A.- 5 J
B.- 10 J
C.- 15 J
D.- 20 J
Solution
ᐃWAB = pᐃV = (10) (2 - 1) = 10 J
WBC = 0 (as V = constant)
From first law of thermodynamics
dQ = dU + dW
dU = 0 (process ABCA is cyclic)
∴ ᐃQ = ᐃWAB + ᐃWBC + ᐃWCA
∴ ᐃWCA = ᐃQ - ᐃWAB - ᐃWBC
= 5 - 10 - 0 = - 5J.
WBC = 0 (as V = constant)
From first law of thermodynamics
dQ = dU + dW
dU = 0 (process ABCA is cyclic)
∴ ᐃQ = ᐃWAB + ᐃWBC + ᐃWCA
∴ ᐃWCA = ᐃQ - ᐃWAB - ᐃWBC
= 5 - 10 - 0 = - 5J.
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