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4 years ago

15

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa

to 4.00×105 Pa. The second process is a compression to a volume of 5.00×10−2 m3 at a constant pressure of 4.00×105 Pa. Find the total work done by the gas during both processes.

1 answer:

Alborosie4 years ago

7 0

<h2>The work done = - 2 x 10⁴ J</h2>

Explanation:

In the first case , the volume is kept constant and pressure varies .

In isothermal process , the work done

W₁ = V x ΔP

here V is the volume of gas and ΔP is the change in pressure

Thus W₁ = 0

Because there is no change in volume , therefore displacement is zero .

In second case pressure is constant , but volume changes

Thus W₂ = P x ΔV

here P is the pressure and ΔV is the change in volume

Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

The total work done W = - 2 x 10⁴ J

Because the work done in compression is negative .

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Para poder solucionar este problema debemos completar el enunciado del problema así como la pregunta, realizando una búsqueda en Internet encontramos el enunciado completo, así como la pregunta.

"Una pelota que se desliza hacia arriba por una pendiente se halla inicialmente a 6 m de la parte más baja de dicha pendiente y tiene una velocidad de 4 m/s. Cinco segundos después se encuentra a 3 m de la parte más baja. Si suponemos una aceleración constante, "

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