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

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(B) For path $iaf$:$Q = 50 \, cal$,$W = 20 \, cal$.According to the first law of thermodynamics,$\Delta U = Q - W$.$\Delta U = 50 - 20 = 30 \, cal$.Si

Vedclass · Accepted Answer

$6, -43, 40 \, cal$

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(B) For path $iaf$:$Q = 50 \, cal$,$W = 20 \, cal$.According to the first law of thermodynamics,$\Delta U = Q - W$.$\Delta U = 50 - 20 = 30 \, cal$.Since internal energy is a state function,$\Delta U$ is the same for any path from $i$ to $f$.$(i)$ For path $ibf$:$Q = 36 \, cal$,$\Delta U = 30 \, cal$.$W = Q - \Delta U = 36 - 30 = 6 \, cal$.$(ii)$ For path $fi$ (reverse path):$W = -13 \, cal$,$\Delta U_{fi} = -\Delta U_{if} = -30 \, cal$.$Q = \Delta U + W = -30 + (-13) = -43 \, cal$.$(iii)$ Given $E_{int,i} = 10 \, cal$:$E_{int,f} = E_{int,i} + \Delta U = 10 + 30 = 40 \, cal$.

Column-$I$	Column-$II$
$(a)$ Adiabatic	$(i)$ $\Delta Q = \Delta U$
$(b)$ Isothermal	$(ii)$ $\Delta Q = \Delta W$
	$(iii)$ $\Delta U = -\Delta W$

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