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 *p*1V1 = *p*2V2** = 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**

**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:**

*p*1 = 0.820 bar

*p*2 = 1.025 bar

V1 = 200 *ml*

V2 = ?

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

*p*1V1 = *p*2V2

v2 = P1V1/P2 = 160 ml

**Example 2**

**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.

*p*1V1 = *p*2V2

*p*1 = 1.2 bar

*p*2 =?

*V*1 = 120 mL

*V*2 = 180 mL

*p*1V1 = *p*2V2

∴ 1.2 × 120 = *p*2 × 180

**p2 =0.8 bar**

**Example 3**

**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:**

*p*1 = 745 mm Hg

*p*2 = ?

V1 = 250 mL

V2 = 200 mL

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

*p*1V1 = *p*2V2

P2 = P1 V1 / V2

**p2 = 931.25 mm Hg**

*The additional pressure required =P2 – P1= 931.25 *– 745

= **186.25 mm.**

**Example 4**

**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,

*p*1 V1 = *p*2 V2

If *p*1 = 1 bar then V1 = 2.27 L

If *p*2 = 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.