worked ExampleA highaltitude balloon contains 60L of helium gas at 206 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 50.0kPa under constant temperature?SolutionApplying Boyle’s Law, P_{1}V_{ 1 }= P_{2}V_{ 2}P_{1 }= 206kPaV_{1}=60LP_{2}= 50kPaV_{2}=?P_{1}V_{ 1 }= P_{2}V_{ 2, }making V_{ 2} the subject of the formula gives;V_{ 2}= P_{1×}V_{ 1 } P_{2}= 206 kPa_{×}60L_{ } 50L= 247.2LFrom the above calculation, a decrease in pressure of the gas at constant temperature leads to an increase in volume. 
Worked ExampleA balloon is inflated in a room at 24^{◦}c has a volume of 4.00L. The balloon is then heated to a temperature of 58^{◦}c. Calculate the new volume of the gas given that temperature is kept constant?SolutionApplying Charles’s Law, V_{1}T_{1} =V_{2}T_{2}T_{1 }= 297KV_{1}=4.00LT_{2 }= 331KV_{2}=?V_{1}T_{1} =V_{2}T_{2} making V_{ 2} the subject of the formula gives;V_{ 2}= T_{2×}V_{ 1 } T_{1}V_{ 2}= 331K_{×}4.00L 297= 4.46L 
At a temperature of 57^{◦}c, nitrogen gas occupies a volume of 750cm^{3. }At what temperature will the gas occupy 100cm^{3}? Express your answer in degrees Celsius.Applying Charles’s Law, V_{1}T_{1} =V_{2}T_{2}T_{1 }=330 KV_{1}=750cm^{3}V_{2 }= 100cm^{3}T_{2}=?V_{1}T_{1} =V_{2}T_{2} making V_{ 2} the subject of the formula gives;T_{ 2}= V_{2×}T_{ 1 }V_{1}T_{ 2}= 100cm3_{×}330K 750cm3= 44273=229^{◦}c 
QuestionThe pressure in an automobile tire is 198kPa at27^{◦}c. Given that the pressure has risen to 225kPa. Calculate the temperate of the air in the tire assuming that the volume is kept constant.Solution:Applying Gay Lussac’s Law, P_{1}T_{1} =P_{2}T_{2}T_{1 }=300 KP_{1}=198kPaP_{2 }= 225kPaT_{2}=?P_{1}T_{1} =P_{2}T_{2} making T_{2} the subject of the formula gives;T_{ 2}= P_{2×}T_{ 1 }P_{1}=225kPa_{×}300K_{ }198kPa= 340.9090 273= 67.9^{◦}c 
Quick question:Briefly explain how combining gas law can be reduced to the other three gas laws.P_{1×}V_{1 }T_{1}=P_{2×}V_{2 }T_{2}The other laws can be obtained from this law by keeping one quantity constant.To illustrate, suppose we hold the temperature constant. Rearranging the combined gas law, we shall obtain the temperature terms on the same side of the equation and then cancelled.
