Class 9 - Unit # 7 : Properties of Matter - Solved Numericals

Boyle's Law

7. The pressure on $9\text{ cm}^3$ of oxygen gas is doubled at a fixed temperature. What will its volume become?

Data:

• Initial Volume ($V_1$) = $9\text{ cm}^3$
• Initial Pressure ($P_1$) = $P_1$
• Final Pressure ($P_2$) = $2P_1$
• Final Volume ($V_2$) = ?

Solution:

According to Boyle’s Law, at a constant temperature, the volume of a given mass of gas is inversely proportional to its pressure:

$$P_1 V_1 = P_2 V_2$$

Rearranging the formula to solve for the final volume ($V_2$):

$$V_2 = \frac{P_1 V_1}{P_2}$$ $$V_2 = \frac{P_1 \times 9\text{ cm}^3}{2P_1}$$ $$V_2 = \frac{9}{2}\text{ cm}^3 = 4.5\text{ cm}^3$$

RESULT: The final volume of the gas is $4.5\text{ cm}^3$.

8. A container holds $30\text{ m}^3$ of air at a pressure of $150,000\text{ Pa}$. If the volume changed to $10\text{ m}^3$ at a constant temperature, what will the pressure of the gas become?

Data:

• Initial Volume ($V_1$) = $30\text{ m}^3$
• Initial Pressure ($P_1$) = $150,000\text{ Pa}$
• Final Volume ($V_2$) = $10\text{ m}^3$
• Final Pressure ($P_2$) = ?

Solution:

According to Boyle’s Law:

$$P_1 V_1 = P_2 V_2$$

Rearranging the formula to isolate the final pressure ($P_2$):

$$P_2 = \frac{P_1 V_1}{V_2}$$ $$P_2 = \frac{150,000\text{ Pa} \times 30\text{ m}^3}{10\text{ m}^3}$$ $$P_2 = 150,000 \times 3 = 450,000\text{ Pa} \quad (\text{or } 4.5 \times 10^5\text{ Pa})$$

RESULT: The final pressure of the gas is $450,000\text{ Pa}$.

9. Air at an atmospheric pressure of $760\text{ mm of Hg}$ is trapped inside a container equipped with a movable piston. When the piston is pulled out slowly so that the volume is increased from $100\text{ dm}^3$ to $150\text{ dm}^3$ at a constant temperature, what will the pressure of the air become?

Data:

• Initial Volume ($V_1$) = $100\text{ dm}^3$
• Initial Pressure ($P_1$) = $760\text{ mm of Hg}$
• Final Volume ($V_2$) = $150\text{ dm}^3$
• Final Pressure ($P_2$) = ?

Solution:

According to Boyle’s Law:

$$P_1 V_1 = P_2 V_2$$

Rearranging the formula to isolate the final pressure ($P_2$):

$$P_2 = \frac{P_1 V_1}{V_2}$$ $$P_2 = \frac{760\text{ mm of Hg} \times 100\text{ dm}^3}{150\text{ dm}^3}$$ $$P_2 = \frac{76,000}{150}\text{ mm of Hg} \approx 506.67\text{ mm of Hg}$$

RESULT: The final pressure of the air is approximately $506.67\text{ mm of Hg}$.