A gas goes from state A to state B via three | vedclass

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(A) According to the first law of thermodynamics, $Q = \Delta U + W$.Since all three paths start at state $A$ and end at state $B$, the change in inte

Vedclass · Accepted Answer

$Q_1 < Q_2 < Q_3$

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(A) According to the first law of thermodynamics, $Q = \Delta U + W$.Since all three paths start at state $A$ and end at state $B$, the change in internal energy $\Delta U$ is the same for all three paths.Therefore, the heat absorbed $Q$ depends only on the work done $W$ by the gas.The work done $W$ by the gas is equal to the area under the $P-V$ curve.From the given diagram, 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, $W_1 Since $Q = \Delta U + W$ and $\Delta U$ is constant, we have $Q_1 < Q_2 < Q_3$.

$A$ gas goes from state $A$ to state $B$ via three different paths as shown in the $P-V$ diagram. If $Q_1, Q_2,$ and $Q_3$ are the heat absorbed by the gas along the three paths, then:

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