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The temperature of an ideal gas in a sealed rigid 0.20-m3 container is reduced from 360 K to270 K, and the final pressure of the gas is 40 kPA. How much work is done by the gas during thisprocess

Respuesta :

Answer:

0 Joule.

Explanation:

Given that the initial volume of the seal rigid container is

[tex]V_{i}=0.20 m^{3}[/tex]

And the temperature changes from 360 K TO 270 K.

And the final pressure is 40 kPA.

Now, this the condition in which the ideal gas in a sealed rigid container which means there is no change in volume.

Si work done is,

[tex]W=PdV[/tex]

There is no change in volume will give zero work done.

Therefore, work is done by gas during this process is 0 J.

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