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

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(B) According to the first law of thermodynamics,for any path,$Q = \Delta U + W$,where $\Delta U$ is the change in internal energy.For path $iaf$:$Q_{

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(B) According to the first law of thermodynamics,for any path,$Q = \Delta U + W$,where $\Delta U$ is the change in internal energy.For path $iaf$:$Q_{iaf} = \Delta U_{iaf} + W_{iaf}$$50 = \Delta U_{iaf} + 20$$\Delta U_{iaf} = 50 - 20 = 30 \, cal$Since internal energy is a state function,the change in internal energy depends only on the initial and final states,not on the path taken.Therefore,$\Delta U_{ibf} = \Delta U_{iaf} = 30 \, cal$.For path $ibf$:$Q_{ibf} = \Delta U_{ibf} + 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 $iaf$,$Q = 50 \, cal$ and $W = 20 \, cal$. Along the path $ibf$,if $Q = 36 \, cal$,find the work $W$ for the path $ibf$ in $cal$.

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