A gas at a pressure of "203 kPa"203 kPa occupies "40.0 L"40.0 L. If the pressure is decreased to "35.0 kPa"35.0 kPa, what is the new volume? Assume the temperature is held constant.

1 Answer
Meave60
Jan 4, 2017

The final volume of helium gas will be "232 L"232 L.

Explanation:

This is an example of Boyle's law, which states that the volume of a given amount of gas, held at a constant temperature, will vary inversely with its pressure. This means that when the pressure is increased, the volume will decrease and vice versa

The equation to use is "P_1V_1=P_2V_2P1V1=P2V2, where PP represents pressure, and VV represents volume.

Known
P_1="203 kPa"P1=203 kPa
V_1="40.0 L"V1=40.0 L
P_2="35.0 kPa"P2=35.0 kPa

Unknown
V_2V2

Solution
Rearrange the equation to isolate V_2V2. Substitute the known values into the equation and solve.

V_2=(P_1V_1)/(P_2)V2=P1V1P2

V_2=(203cancel"kPa"xx40.0"L")/(35.0cancel"kPa")="232 L"

The final volume of helium gas will be "232 L".

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