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Air that initially occupies 0.25 m3 at a gauge pressure of 110 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

Respuesta :

Answer:

W = 234 J

Explanation:

Initial volume of air is given as

[tex]V_1 = 0.25 m^3[/tex]

[tex]P_1 = 110 kPa[/tex]

finally it is expanded isothermally to new pressure of 101.3 kPa

[tex]P_2 = 101.3 kPa[/tex]

now by isothermal expansion we know

[tex]P_1V_1 = P_2V_2[/tex]

[tex](110 kPa)(0.25) = (101.3)V_2[/tex]

[tex]V_2 = 0.27 m^3[/tex]

So work done in above process is given as

[tex]W_1 = P_1V_1ln(\frac{P_1}{P_2})[/tex]

[tex]W_1 = (110\times 10^3)(0.25) ln(\frac{110}{101.3})[/tex]

[tex]W_1 = 2.26 \times 10^3 J[/tex]

Now it is cooled to initial volume at constant pressure

so here in this case work done is given as

[tex]W_2 = P(V_2 - V_1)[/tex]

[tex]W_2 = (101.3 \times 10^3)(0.25 - 0.27)[/tex]

[tex]W_2 = -2026 J[/tex]

Now total work done is given as

[tex]W = W_1 + W_2[/tex]

[tex]W = 2260 - 2026 = 234 J[/tex]

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