An expandable container of oxygen has a volume of 30.0mL at a pressure of 36.7psi. If the pressure of the oxygen is reduced to 25.0psi what would the new volume of the contain be

44.04mL

Explanation:Parameters given

V1 = 30.0mL

P1 = 36.7psi

P2 = 25.0psi

V2 = ??

From Boyle's gas law, which states that "the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature"

This means that,

the pressure of a gas tends to increase as the volume of the container decreases, and also the pressure of a gas tends to decrease as the volume of the container increases.

Mathematically, Boyle's can be represented as shown below

P= k/V

Where P = Pressure, V = Volume and k is constant

Therefore,

PV = k

P1V1 = P2V2 =PnVn

Using the formula

P1V1 = P2V2

V2 = P1V1/P2

V2 = (36.7psi × 30.0mL) / 25.0psi

V2 = 1101.0/25.0

V2 = 44.04mL

44.04 mL

Explanation:

From Boyle's law,

PV = P'V' Equation 1

Where P = initial pressure, V = initial volume, P' = Final pressure, V' = final volume.

make V' the subject of the equation

V' = PV/P' Equation 2

Given: P = 36.7 psi, V == 30 mL = (30/1000) L = 0.03 L, P' = 25 psi.

Substitute into equation 2

V' = 36.7(0.03)/25

V' = 1.101/25

V' = 0.04404 L

V' = 44.04 mL.

Hence the new volume = 44.04 mL

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