# Boyle’s law (constant temperature) – problems and solutions

1. Some ideal gases initially have pressure P and volume V. If the gas undergoes isothermal process so that the final pressure becomes 4 times the initial pressure, then the final volume of gas is…

__Advertisement
__

Initial pressure (P_{1}) = P

Final pressure (P_{2}) = 4P

Initial volume (V_{1}) = V

__Wanted:__ Final volume (V_{2})

__Solution :__

The formula of Boyle’s law :

P V = constant

P_{1} V_{1} = P_{2} V_{2}

(P)(V) = (4P)(V_{2})

V = 4 V_{2}

V_{2 }= V / 4 = ¼ V

The final volume of gases is ¼ times the initial volume.

2. In a closed container, the gas expands so that the final volume becomes 2 times the initial volume (V = initial volume, P = initial pressure).

__Known :__

Initial pressure (P_{1}) = P

Initial volume (V_{1}) = V

Final volume (V_{2}) = 2V

__Wanted ____:__ Final pressure (P_{2})

__Solution :__

P_{1} V_{1} = P_{2} V_{2}

P V = P_{2} (2V)

P = P_{2} (2)

P_{2 }= P / 2 = ½ P

The gases pressure becomes ½ times the initial pressure.

3. In a closed container, gases having a pressure of 2 atm and a volume of 1 liter. If gas pressure becomes 4 atm then gas volume becomes …

__Known :__

Initial pressure (P_{1}) = 2 atm = 2 x 10^{5} Pa

Final pressure (P_{2}) = 4 atm = 4 x 10^{5} Pa

Initial volume (V_{1}) = 1 liter = 1 dm^{3} = 1 x 10^{-3} m^{3}

__Wanted__ : Final volume (V_{2})

__Solution :__

P_{1} V_{1} = P_{2} V_{2}

(2 x 10^{5})(1 x 10^{-3}) = (4 x 10^{5}) V_{2}

(1)(1 x 10^{-3}) = (2) V_{2}

1 x 10^{-3} = (2) V_{2}

V_{2 }= ½ x 10^{-3}

V_{2} = 0.5 x 10^{-3 }m^{3} = 0.5 dm^{3} = 0.5 liters