An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure. If net heat supplied to the gas in the cycle is $5 J$ , the work done by the gas in the process $C \to A$ is :

+ 5 J

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+ 10 J

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+ 15 J

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+ 20 J

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Solution

The correct option is D $+ 15 J$
Internal energy is a state function. Hence, its value at point A is constant before and after the cyclic process.
Also, $Δ U = q + w = 0$
Thus, $w = - q$
$W (A C) = W (A B) + W (B C) - q$ ......(1)
$W (A B) = P Δ V = 10 \times 2 = 20 J$ ......(2)
$W B C) = P Δ V = 0 Δ V = 0$ ......(3)
Substittue (2) and (3) in (1)
$W (A C) = 20 - 5 = 15 J$

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An ideal gas is taken through the cycle A→B→C→A, as shown in the figure. If net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process C→A is :

The correct option is D +15JInternal energy is a state function. Hence, its value at point A is constant before and after the cyclic process.Also, ΔU=q+w=0 Thus, w=−qW(AC)=W(AB)+W(BC)−q......(1)W(AB)=PΔV=10×2=20J......(2)WBC)=PΔV=0ΔV=0......(3)Substittue (2) and (3) in (1)W(AC)=20−5=15J

An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure. If net heat supplied to the gas in the cycle is $5 J$ , the work done by the gas in the process $C \to A$ is :