Boyles law formula, relationship, example problem

1 answer

Boyle's Law Formula

Boyle's Law states that the pressure of a given mass of gas is inversely proportional to its volume when the temperature is held constant. The formula is expressed as:

\[ P_1 V_1 = P_2 V_2 \]

Where:

  • \( P_1 \) = initial pressure
  • \( V_1 \) = initial volume
  • \( P_2 \) = final pressure
  • \( V_2 \) = final volume

Relationship

From Boyle's Law, we can derive the relationship:

  • If the volume of a gas decreases (\( V \downarrow \)), the pressure increases (\( P \uparrow \)), assuming temperature remains constant.
  • If the volume of a gas increases (\( V \uparrow \)), the pressure decreases (\( P \downarrow \)), assuming temperature remains constant.

This means that the product of pressure and volume for a fixed amount of gas at constant temperature is a constant value.

Example Problem

Problem: A balloon has a volume of 2.0 L at a pressure of 1.0 atm. If the gas is compressed to a volume of 1.0 L, what will the new pressure be?

Given:

  • \( V_1 = 2.0 , \text{L} \)
  • \( P_1 = 1.0 , \text{atm} \)
  • \( V_2 = 1.0 , \text{L} \)

Find:

  • \( P_2 \)

Using Boyle’s Law, we can rearrange the formula to solve for \( P_2 \):

\[ P_2 = \frac{P_1 V_1}{V_2} \]

Substituting the known values:

\[ P_2 = \frac{1.0 , \text{atm} \times 2.0 , \text{L}}{1.0 , \text{L}} \]

\[ P_2 = \frac{2.0 , \text{atm} \cdot \text{L}}{1.0 , \text{L}} = 2.0 , \text{atm} \]

Conclusion

Thus, if the balloon is compressed to a volume of 1.0 L, the new pressure will be 2.0 atm.