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A sealed container filled with argon gas at 35C has a pressure of 832 torr. If the volume of the container is decreased by a factor of two, what will happen to the pressure?

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Answer:

Pressure is increased by a factor of 2

Step-by-step explanation:


P_(1)=832 torr and
P_(2) is unknown

where
P_(1) is initial pressure and
P_(2) is the final pressure


V1 = V,
V2 = V/2 where
V_(1) and
V_(2) are the initial and final volumes respectively

From gas laws


P_(1)V_(1) = P_(2)V_(2)


832* V = P_(2)* \frac {V}{2}

Pressure
P_(2) = 1664 torr

The above pressure of 1664 torr is double the pressure of 832 torr. Therefore, we conclude that the pressure is increased by a factor of 2

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