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

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(A) For all three processes,the initial state $A$ and final state $B$ are the same.Since internal energy $U$ is a state function,the change in interna

Vedclass · Accepted Answer

${Q_1} < {Q_2} < {Q_3}$

Vedclass · Answer

(A) For all three processes,the initial state $A$ and final state $B$ are the same.Since internal energy $U$ is a state function,the change in internal energy $\Delta U = U_B - U_A$ is the same for all three paths.According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta W$ is the work done by the gas.In a $P-V$ diagram,the work done $\Delta W$ is equal to the area under the curve with the volume axis.From the figure,the area under the curves is such that $(Area)_1 Since $\Delta U$ is constant,$\Delta Q$ is directly proportional to $\Delta W$.Therefore,${Q_1} < {Q_2} < {Q_3}$.

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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