Question

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

Answer 1

Answer:

0 Joule.

Explanation:

Given that the initial volume of the seal rigid container is

$V_{i}=0.20 m^{3}$

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,

$W=PdV$

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

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