When a system is taken from state i to state | vedclass

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(A) According to the first law of thermodynamics,the heat supplied to a system is equal to the sum of the change in internal energy and the work done

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$6$

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(A) According to the first law of thermodynamics,the heat supplied to a system is equal to the sum of the change in internal energy and the work done by the system: $Q = \Delta U + W$.For the path $iaf$:$Q_{iaf} = 50 \, cal$$W_{iaf} = 20 \, cal$Using the first law: $50 = \Delta U + 20$Therefore,the change in internal energy between states $i$ and $f$ is $\Delta U = 30 \, cal$.Since internal energy is a state function,it depends only on the initial and final states and not on the path taken. Thus,for path $ibf$,the change in internal energy remains $\Delta U = 30 \, cal$.For the path $ibf$:$Q_{ibf} = 36 \, cal$Using the first law: $Q_{ibf} = \Delta U + W_{ibf}$$36 = 30 + W_{ibf}$$W_{ibf} = 36 - 30 = 6 \, cal$.

When a system is taken from state $i$ to state $f$ along the path $iaf,$ it is found that $Q = 50 \, cal$ and $W = 20 \, cal$. Along the path $ibf$,$Q = 36 \, cal$. Work done along the path $ibf$ will be ........... $cal$.

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