High School Physics + more

# Solving Numerical problems using Boyle’s law

Last updated on October 23rd, 2021 at 10:23 am

Here we will present a set of solved numerical problems that use Boyle’s law formula to solve the numerical problems on gas.

The primary formula that will be used to solve these numerical problems is p1V1 = p2V2 = constant. where p1 and V1 are initial pressure and initial volume of the gas respectively. Similarly, p2 and V2 are the final pressure and final volume of the gas respectively.

## Numerical problems based on Boyle’s law – with solution

### Example 1

A gas occupies 200 mL at a pressure of 0.820 bar at 20°C. How much volume will it occupy when it is subjected to external pressure of 1.025 bar at the same temperature?

Solution:

p1 = 0.820 bar

p2 = 1.025 bar

V1 = 200 ml

V2 = ?

Since temperature is constant, therefore, by applying Boyle’s law,

p1V1 = p2V2

v2 = P1V1/P2 = 160 ml

### Example 2

A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?

Solution: Since temperature and amount of gas remain constant, therefore, Boyle’s law is applicable.

p1V1 = p2V2

p1 = 1.2 bar

p2 =?

V1 = 120 mL

V2 = 180 mL

p1V1 = p2V2

∴ 1.2 × 120 = p2 × 180

p2 =0.8 bar

### Example 3

A gas occupies a volume of 250 mL at 745 mm Hg and 25°C. What additional pressure is required to reduce the gas volume to 200 mL at the same temperature?

Solution:

p1 = 745 mm Hg

p2 = ?

V1 = 250 mL

V2 = 200 mL

Since temperature remains constant, therefore, by applying Boyle’s law,

p1V1 = p2V2

P2 = P1 V1 / V2

p2 = 931.25 mm Hg

The additional pressure required =P2 P1= 931.25 – 745

= 186.25 mm.

### Example 4

A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, up to what volume can the balloon be expanded?

Solution: According to Boyle’s law, at constant temperature,

p1 V1 = p2 V2

If p1 = 1 bar then V1 = 2.27 L

If p2 = 0.2 bar, then

V2 = p1v1/p2= 1 bar x 2.27 L/0.2 bar = 11.35 L

Since the balloon bursts at 0.2 bar pressure, the volume of the balloon should be less than 11.35 L.