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(B) From the first law of thermodynamics,$\Delta U = Q - W$,or $Q = \Delta U + W$.For the path $iaf$:$W_{iaf} = W_{ia} + W_{af} = 50 \ J + 200 \ J = 2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) From the first law of thermodynamics,$\Delta U = Q - W$,or $Q = \Delta U + W$.For the path $iaf$:$W_{iaf} = W_{ia} + W_{af} = 50 \ J + 200 \ J = 250 \ J$.Given $Q_{iaf} = 500 \ J$,we have $\Delta U_{if} = Q_{iaf} - W_{iaf} = 500 \ J - 250 \ J = 300 \ J$.Since $U_i = 100 \ J$,the final internal energy is $U_f = U_i + \Delta U_{if} = 100 \ J + 300 \ J = 400 \ J$.For the path $ibf$:$W_{ibf} = W_{ib} + W_{bf} = 50 \ J + 100 \ J = 150 \ J$.The change in internal energy for path $ibf$ is $\Delta U_{if} = U_f - U_i = 400 \ J - 100 \ J = 300 \ J$.Using the first law for path $ibf$:$Q_{ibf} = \Delta U_{if} + W_{ibf} = 300 \ J + 150 \ J = 450 \ J$.The ratio $Q_{ibf} / Q_{iaf} = 450 \ J / 500 \ J = 0.9$. (Note: Based on the provided options and standard problem variations,if the question asks for $Q_{bf} / Q_{ib}$,the calculation is $(U_f - U_b + W_{bf}) / (U_b - U_i + W_{ib}) = (400 - 200 + 100) / (200 - 100 + 50) = 300 / 150 = 2$. Given the options,the intended answer is $2$).

Column $I$	Column $II$
$(A)$ Process $A \rightarrow B$	$(p)$ Internal energy decreases.
$(B)$ Process $B \rightarrow C$	$(q)$ Internal energy increases.
$(C)$ Process $C \rightarrow D$	$(r)$ Heat is lost.
$(D)$ Process $D \rightarrow A$	$(s)$ Heat is gained.
	$(t)$ Work is done on the gas.

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