A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 232.kPa. The piston is now pulled up, expanding the gas, until the gas has a final volume of 29.0L. Calculate the final pressure of the gas. Be sure your answer has the correct number of significant digits.

The final pressure of the gas is 80.0kPa

Explanation:

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"

Mathematically, P ∝ (1/V)

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.

Since P ∝ (1/V), we can then write that

P = k(1/V)

Where P is the pressure, V is the volume and k is the proportionality constant

Therefore,

PV = k

We can then write that

P1V1 = P2V2 = P3V3 = ...

Hence, P1V1 = P2V2

Where P1 is the initial pressure of the gas

P2 is the final pressure of the gas

V1 is the initial volume of the gas

and V2 is the final volume of the gas

From the question,

V1 = 10.0L

P1 = 232.0 kPa

V2 = 29.0L

P2 = ??

From P1V1 = P2V2

232.0kPa × 10.0L = P2 × 29.0L

Hence,

P2 = (232.0kPa × 10.0L) / 29.0L

P2 = 80.0kPa

Hence, the final pressure of the gas is 80.0kPa

80 kPa or 8.0×10⁴ Pa.

Explanation:

From Boyle's law,

PV = P'V' Equation 1

Where P = Initial pressure, V = Initial volume, P' = Final pressure, V' = Final volume.

Make P' the subject of the equation

P' = PV/V' Equation 2

Given: P = 232 kPa, V = 10.0 L, V' = 29.0 L

Substitute into equation 2

P' = 232(10)/29

P' = 80 kPa or 8.0×10⁴ Pa.

Hence the final pressure of the gas = 80 kPa or 8.0×10⁴ Pa.

volume: 22

density: 7.04

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