An ideal gas is taken from point A to point C | vedclass

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Question

(A) Since internal energy $(\Delta U)$ is a state function,the change in internal energy is the same for both processes $AOC$ and $ABC$,i.e.,$\Delta U

Vedclass · Accepted Answer

Process $AOC$ requires more heat than process $ABC$.

Vedclass · Answer

(A) Since internal energy $(\Delta U)$ is a state function,the change in internal energy is the same for both processes $AOC$ and $ABC$,i.e.,$\Delta U_{AOC} = \Delta U_{ABC}$.According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,where $W$ is the work done.The work done $W$ is equal to the area under the $P-V$ curve.From the figure,the area under the curve $AOC$ is greater than the area under the path $ABC$ (which is a rectangle $V_0BC2V_0$ plus the area under $AB$). Specifically,the area under $AOC$ covers the entire region under the curve $AOC$,while the area under $ABC$ is only the area under the line $BC$ (since $AB$ is a vertical line,the area under $AB$ is zero).Therefore,$W_{AOC} > W_{ABC}$.Since $\Delta U_{AOC} = \Delta U_{ABC}$ and $W_{AOC} > W_{ABC}$,it follows that $\Delta Q_{AOC} > \Delta Q_{ABC}$.

An ideal gas is taken from point $A$ to point $C$ on a $P-V$ diagram through two processes $AOC$ and $ABC$ as shown in the figure. Process $AOC$ is isothermal.

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