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An ideal gas described by Ti=291K, Pi=1.50bar, and Vi=13.3L is heated at constant volume until P=15.0bar. It then undergoes a reversible isothermal expansion until P=1.50bar. It is then restored to its original state by the extraction of heat at constant pressure. Calculate w for step 2 (P, Vi, T → Pi, V2, T).

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sebassandin

Answer:

[tex]W=-4601.4J[/tex]

Explanation:

Hello,

In this case, the steps are:

[tex]291K,1.50bar, 13.3L \rightarrow 15.0bar 13.3L,T_2\rightarrow T_2, V_2, 1.50bar\rightarrow 291K,1.50bar, 13.3L[/tex]

In such a way, the work per mole (w) for that isothermal process turns out:

[tex]w=RTln(\frac{P_1}{P_2} )=8.314\frac{J}{mol*K}*291K*\frac{1.50bar}{15.0bar} \\\\w=-5570.8\frac{J}{mol}[/tex]

In addition, if the moles are required, since it is an ideal gas:

[tex]n=\frac{PV}{RT}=\frac{1.50bar*13.3L}{0.083\frac{bar*L}{mol*K}*291K} =0.826mol[/tex]

So the work is:

[tex]W=-5570.8\frac{J}{mol} *0.826mol=-4601.4J[/tex]

Best regards.

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