Boyle's law developed by Robert Boyle in 1662, states that if we keep the temperature of a gas constant in a sealed container. Its pressure (P) varies inversely with its volume (V). In other words at any given temperature if we pressurize a gas its volume will be reduced proportionately to the pressure change. If we increase the volume of a gas its pressure will increase.

**"For a fixed mass of ideal gas at fixed temperature the product of pressure and volume is constant".**

**Boyle's Law Formula** is expressed symbolically as

- P is pressure of the gas
- V is volume of the gas
- k is a constant, and has units of force times distance

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Solved problems based on Boyle's law are given below. ### Solved Examples

**Question 1: **A sample of gaseous nitrogen in a 65.0 L automobile air bag has a pressure of 745 mm Hg. If this sample is transferred to a 25.0 L bag at the same temperature. what is the pressure of the gas in the 25.0 L bag?

** Solution: **

It is often useful to make a table of the information provided.

We know that P_{1}V_{1} = P_{2}V_{2}

Therefore,

P_{1} = $\frac{P_{1}V_{1}}{V_{2}}$

P_{1} = $\frac{[745\ mm\ Hg][65.0\ L]}{25.0\ L}$

P_{1} = 1940 mm Hg

**Question 2: **A sample of neon (Ne) occupies 4.00L at a pressure of 5.00 $\times$ 10^{4} Pa and a temperature of 273K. Determine the volume of the sample at STP?

** Solution: **

The initial conditions are

P_{1} = 5.00 $\times$ 10^{4} Pa, V_{1} = 4.00L

The final conditions are

P_{2} = 101325Pa, V_{2} = unknown

Rearrange

to solve for V_{2}

V_{2} = $\frac{P_{1}V_{1}}{P_{2}}$

V_{2 }= $\frac{[5.00 \times 10^{4} Pa][4.00\ L]}{1.01325 \times 10^{5} Pa}$

V_{2 }= $\frac{2.00 \times 10^{5}}{1.01325 \times 10^{5}}$ L

V_{2 }= 1.97L

It is often useful to make a table of the information provided.

Initial conditions |
Final conditions |

P_{1} = 745 mm Hg |
P_{2} = ? |

V_{1} = 65.0 L |
V_{2} = 25.0L |

We know that P

Therefore,

P

P

P

The initial conditions are

P

The final conditions are

P

Rearrange

to solve for V

V

V

V

V