Question

A gas is compressed from an initial volume of $0.42 \mathrm{m}^{3}$ to a final volume of $0.12 \mathrm{m}^{3} .$ During the quasi-equilibrium process, the pressure changes with volume according to the relation $P=a V+b,$ where $a=-1200 \mathrm{kPa} / \mathrm{m}^{3}$ and $b=600 \mathrm{kPa}$ Calculate the work done during this process ( $a$ ) by plotting the process on a $P$ - $V$ diagram and finding the area under the process curve and ( $b$ ) by performing the necessary integrations.

          A gas is compressed from an initial volume of $0.42 \mathrm{m}^{3}$ to a final volume of $0.12 \mathrm{m}^{3} .$ During the quasi-equilibrium process, the pressure changes with volume according to the relation $P=a V+b,$ where $a=-1200 \mathrm{kPa} / \mathrm{m}^{3}$ and $b=600 \mathrm{kPa}$
Calculate the work done during this process ( $a$ ) by plotting the process on a $P$ - $V$ diagram and finding the area under the process curve and ( $b$ ) by performing the necessary integrations.

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Added by Antonio G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/26/2023

Step 1

First, we need to find the initial and final pressures using the given relation $P = aV + b$. Initial pressure, $P_i = aV_i + b = -1200(0.42) + 600 = -504 + 600 = 96\,\text{kPa}$ Final pressure, $P_f = aV_f + b = -1200(0.12) + 600 = -144 + 600 = 456\,\text{kPa}$ Show more…

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A gas is compressed from an initial volume of $0.42 \mathrm{m}^{3}$ to a final volume of $0.12 \mathrm{m}^{3} .$ During the quasi-equilibrium process, the pressure changes with volume according to the relation $P=a V+b,$ where $a=-1200 \mathrm{kPa} / \mathrm{m}^{3}$ and $b=600 \mathrm{kPa}$ Calculate the work done during this process ( $a$ ) by plotting the process on a $P$ - $V$ diagram and finding the area under the process curve and ( $b$ ) by performing the necessary integrations.

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