Question

Difficulty: MediumBoyle's Law and Pressure-Volume Relationship

A sample of gas is enclosed in a cylinder fitted with a frictionless piston at a constant temperature of 27C27^\circ\text{C}. The initial pressure and volume of the gas are 1.2×105 N/m21.2 \times 10^5\text{ N/m}^2 and 600 cm3600\text{ cm}^3 respectively. If the pressure on the gas is increased to 3.6×105 N/m23.6 \times 10^5\text{ N/m}^2, what is the final volume of the gas?

  1. 200 cm3200\text{ cm}^3Answer
  2. B
    1800 cm31800\text{ cm}^3
  3. C
    5400 cm35400\text{ cm}^3
  4. D
    432 cm3432\text{ cm}^3

Answer

The final volume of the gas is 200 cm3200\text{ cm}^3.
According to Boyle's Law, the pressure and volume of a fixed mass of gas are inversely proportional at constant temperature (P1V1=P2V2P_1V_1 = P_2V_2). Substituting the given values gives (1.2×105 N/m2)(600 cm3)=(3.6×105 N/m2)V2(1.2 \times 10^5\text{ N/m}^2)(600\text{ cm}^3) = (3.6 \times 10^5\text{ N/m}^2)V_2, which simplifies to V2=200 cm3V_2 = 200\text{ cm}^3.

Step-by-Step Solution

1
Identify the given quantities and state the relevant gas law formula.
P1=1.2×105 N/m2P_1 = 1.2 \times 10^5\text{ N/m}^2, V1=600 cm3V_1 = 600\text{ cm}^3, P2=3.6×105 N/m2P_2 = 3.6 \times 10^5\text{ N/m}^2. Since temperature is constant, apply Boyle's Law: P1V1=P2V2P_1V_1 = P_2V_2.
Boyle's Law states that for a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure.
2
Rearrange the equation to solve for the unknown final volume (V2V_2).
V2=P1V1P2V_2 = \frac{P_1 V_1}{P_2}
Isolating V2V_2 allows direct substitution of the known variables.
3
Substitute the values into the rearranged equation and compute the result.
V2=(1.2×105 N/m2)×(600 cm3)3.6×105 N/m2=7.2×1073.6×105=200 cm3V_2 = \frac{(1.2 \times 10^5\text{ N/m}^2) \times (600\text{ cm}^3)}{3.6 \times 10^5\text{ N/m}^2} = \frac{7.2 \times 10^7}{3.6 \times 10^5} = 200\text{ cm}^3
Carrying out the arithmetic yields the compressed gas volume.

Key Concept

Boyle's Law (P1V1=P2V2P_1V_1 = P_2V_2 at constant temperature)
Rate this question