An ideal gas of mass m in a state A goes to a | vedclass

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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Since the initial state $A$ and final state $B$ are the same for all

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$Q_{1} < Q_{2} < Q_{3}$

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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Since the initial state $A$ and final state $B$ are the same for all three processes,the change in internal energy $\Delta U$ is the same for all paths.Therefore,$\Delta Q = \Delta U + \Delta W$ implies that the heat absorbed $\Delta Q$ is directly proportional to the work done $\Delta W$ (since $\Delta U$ is constant).The work done $\Delta W$ in a $p-V$ diagram is equal to the area under the curve.From the figure,the area under path $1$ is the smallest,the area under path $2$ is intermediate,and the area under path $3$ is the largest.Thus,$(	ext{Area})_{1} Consequently,$Q_{1} < Q_{2} < Q_{3}$.

An ideal gas of mass $m$ in a state $A$ goes to another state $B$ via three different processes as shown in the figure. If $Q_{1}, Q_{2}$,and $Q_{3}$ denote the heat absorbed by the gas along the three paths,then

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