Examkin

Question

Difficulty: HardBoyle's Law and Pressure-Volume Relationship

A trapped gas sample has a volume of 4.0 dm34.0\text{ dm}^3 at a pressure of 570 mmHg570\text{ mmHg} under isothermal conditions. If the pressure is altered to 1.5 atm1.5\text{ atm}, what is the final volume occupied by the gas? (1 atm=760 mmHg1\text{ atm} = 760\text{ mmHg})

  1. 2.0 dm32.0\text{ dm}^3Answer
  2. B
    8.0 dm38.0\text{ dm}^3
  3. C
    1520.0 dm31520.0\text{ dm}^3
  4. D
    1.52 dm31.52\text{ dm}^3

Answer

The final volume occupied by the gas is 2.0 dm32.0\text{ dm}^3.
According to Boyle's Law, the pressure and volume of a fixed mass of gas at constant temperature are inversely proportional (P1V1=P2V2P_1 V_1 = P_2 V_2). First, convert 570 mmHg570\text{ mmHg} into atmospheres: 570760=0.75 atm\frac{570}{760} = 0.75\text{ atm}. Substituting the values gives 0.75 atm×4.0 dm3=1.5 atm×V20.75\text{ atm} \times 4.0\text{ dm}^3 = 1.5\text{ atm} \times V_2, which yields V2=2.0 dm3V_2 = 2.0\text{ dm}^3.

Step-by-Step Solution

1
Convert the initial pressure P1P_1 from mmHg\text{mmHg} to atm\text{atm} to match the unit of P2P_2.
P1=570 mmHg760 mmHg/atm=0.75 atmP_1 = \frac{570\text{ mmHg}}{760\text{ mmHg/atm}} = 0.75\text{ atm}
Units of pressure must be consistent before substituting into gas law equations.
2
Apply Boyle's Law formula for an isothermal process (P1V1=P2V2P_1 V_1 = P_2 V_2).
0.75 atm×4.0 dm3=1.5 atm×V20.75\text{ atm} \times 4.0\text{ dm}^3 = 1.5\text{ atm} \times V_2
Boyle's Law states that for a fixed mass of gas at constant temperature, pressure and volume are inversely proportional.
3
Solve the algebraic equation for the final volume V2V_2.
V2=0.75×4.01.5=2.0 dm3V_2 = \frac{0.75 \times 4.0}{1.5} = 2.0\text{ dm}^3
Dividing both sides by 1.5 atm1.5\text{ atm} isolates the unknown final volume.

Key Concept

Boyle's Law (P1V1=P2V2P_1 V_1 = P_2 V_2 at constant temperature)
Estimated Time:2m 0s
Rate this question

Topics

Boyle's Law and Pressure-Volume Relationship

More questions from this topic