A gas with a volume of 4.0 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?

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A gas with a volume of 4.0 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?

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Answer 1 · 2017-04-01T10:07:43

We use old $Boyle's Law$ $"Boyle's Law"$ , $P V = k,$ $PV=k,$ where $k$ $k$ is a $constant$ $"constant"$ .

Explanation:

And the utility of this law means that we could use non-standard units, if our units are consistent.

Given that $P V = k,$ $PV=k,$ (temperature constant), it follows that...........

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ ..........

We solve for $P_2=(P_1V_1)/(V_2)=(205*kPaxx4.0*cancelL)/(12.0*cancelL)$

$~=68*kPa$

The pressure has reasonably decreased............Why $"reasonably?"$ And given consistent units, I could use $"pints"$ , $"foot lbs"$ , $"atmospheres"$ whatever...........

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A gas with a volume of 4.0 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?

1 Answer

Explanation:

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