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A certain amount of gas is sealed in a container of which the volume can change. The relationship between the pressure and volume of the gas at 20 °C is investigated. The results of the experiment are given in the table below.

Physics

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3 parts
5.1

Boyle’s Law\text{Boyle's Law}

5.4

81.2 dm3\text{81.2 dm}^3

5.8

20.0 mol\text{20.0 mol}

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Part 5

Part 5.1

Boyle’s Law\text{Boyle's Law}

The gas law represented by the results of the experiment, which shows an inverse relationship between pressure and volume at constant temperature, is Boyle's Law.

Boyle’s Law\boxed{\text{Boyle's Law}}

Part 5.2

A hypothesis for the investigation could be:
If the pressure of a gas is increased at a constant temperature, then its volume will decrease proportionally.

Part 5.3

To draw a graph of volume versus pressure, you would plot the pressure values (kPa) on the x-axis and the corresponding volume values (dm3\text{dm}^3) on the y-axis. The points would form a curve that decreases as pressure increases, representing an inverse relationship.

Part 5.4

81.2 dm3\text{81.2 dm}^3

  1. State Boyle's Law.

    Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional: P1V1=P2V2P_1V_1 = P_2V_2.

  2. Select an initial data point from the table.

    Using the first data point:
    P1=140 kPaP_1 = 140 \text{ kPa}
    V1=348 dm3V_1 = 348 \text{ dm}^3

  3. Use the given final pressure to calculate the final volume.

    P2=600 kPaP_2 = 600 \text{ kPa}

    V2=P1V1P2V_2 = \frac{P_1V_1}{P_2} V2=140 kPa×348 dm3600 kPaV_2 = \frac{140 \text{ kPa} \times 348 \text{ dm}^3}{600 \text{ kPa}} V2=48720 kPa⋅dm3600 kPaV_2 = \frac{48720 \text{ kPa} \cdot \text{dm}^3}{600 \text{ kPa}} V2=81.2 dm3V_2 = 81.2 \text{ dm}^3

    The calculated volume of the gas at 600 kPa600 \text{ kPa} is 81.2 dm3\boxed{\text{81.2 dm}^3}.

Part 5.5

The calculated volume in Question 5.4 assumes ideal gas behavior, where gas particles have no volume and no intermolecular forces. Real gases, however, do have finite particle volume and experience intermolecular forces. At high pressures (like 600 kPa600 \text{ kPa}), the volume occupied by the gas particles themselves becomes significant, and intermolecular forces become more pronounced, causing the real gas to occupy a slightly larger volume than predicted by ideal gas laws.

Part 5.6

A gas will deviate from ideal behavior at LOW temperature conditions.

Part 5.7

At low temperatures, the kinetic energy of gas particles decreases, making the attractive intermolecular forces between them more significant. These forces cause the particles to be closer together and reduce the effective pressure they exert, leading to deviations from ideal gas behavior.

Part 5.8

20.0 mol\text{20.0 mol}

  1. Convert the temperature to Kelvin.

    T=20 °C+273.15=293.15 KT = 20 \text{ °C} + 273.15 = 293.15 \text{ K}
  2. Select an initial pressure and volume from the table.

    Using the first data point:
    P=140 kPaP = 140 \text{ kPa}
    V=348 dm3V = 348 \text{ dm}^3

  3. Use the Ideal Gas Law (PV=nRTPV = nRT) to calculate the number of moles (nn).

    The ideal gas constant R=8.314 kPa⋅dm3/(mol⋅K)R = 8.314 \text{ kPa} \cdot \text{dm}^3 \text{/(mol} \cdot \text{K)}.

    n=PVRTn = \frac{PV}{RT} n=140 kPa×348 dm38.314 kPa⋅dm3/(mol⋅K)×293.15 Kn = \frac{140 \text{ kPa} \times 348 \text{ dm}^3}{8.314 \text{ kPa} \cdot \text{dm}^3 \text{/(mol} \cdot \text{K)} \times 293.15 \text{ K}} n=487202437.991 moln = \frac{48720}{2437.991} \text{ mol} n=19.983 moln = 19.983 \text{ mol}

    Rounding to three significant figures:

    n=20.0 moln = 20.0 \text{ mol}

    The number of moles of the gas in the container is 20.0 mol\boxed{\text{20.0 mol}}.

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