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A group of learners perform an experiment to verify a gas law. The results of the experiment are shown in the graph of pressure vs the inverse of volume drawn below.

Physics

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3 parts
5.1

Boyle’s Law\text{Boyle's Law}

5.4

20 cm320 \text{ cm}^3

5.5

360 kPa360 \text{ kPa}

Step-by-step solution

6 steps

Part 5.1

The graph shows a direct proportionality between pressure (PP) and the inverse of volume (1V\frac{1}{V}). This relationship is described by Boyle's Law.

Boyle’s Law\boxed{\text{Boyle's Law}}

Part 5.2

An investigative question for this experiment could be:
"How does the pressure of a gas change as its volume is varied, while keeping the temperature and amount of gas constant?"

Part 5.3

The two variables that must be kept constant during the experiment to verify Boyle's Law are:

  • The temperature of the gas.
  • The amount (number of moles) of the gas.

Part 5.4

20 cm320 \text{ cm}^3

  1. Locate the pressure value on the y-axis.

    From the graph, when the pressure is 200 kPa200 \text{ kPa}, find the corresponding value on the x-axis.

  2. Read the inverse of volume from the graph.

    At P=200 kPaP = 200 \text{ kPa}, the value on the x-axis is 5×10−2 cm−35 \times 10^{-2} \text{ cm}^{-3}.
    So, 1volume=5×10−2 cm−3\frac{1}{\text{volume}} = 5 \times 10^{-2} \text{ cm}^{-3}.

  3. Calculate the volume.

    Volume=15×10−2 cm−3=10.05 cm−3=20 cm3\text{Volume} = \frac{1}{5 \times 10^{-2} \text{ cm}^{-3}} = \frac{1}{0.05 \text{ cm}^{-3}} = 20 \text{ cm}^3

    The volume occupied by the gas is 20 cm3\boxed{20 \text{ cm}^3}.

Part 5.5

360 kPa360 \text{ kPa}

  1. Calculate the inverse of the given volume.

    Given volume V=11.11 cm3V = 11.11 \text{ cm}^3.

    1V=111.11 cm3≈0.090009 cm−3\frac{1}{V} = \frac{1}{11.11 \text{ cm}^3} \approx 0.090009 \text{ cm}^{-3}

  2. Determine the constant of proportionality (slope) from the graph.

    The graph is a straight line passing through the origin. Using the point (5×10−2 cm−3,200 kPa)(5 \times 10^{-2} \text{ cm}^{-3}, 200 \text{ kPa}):

    P=k(1V)P = k \left(\frac{1}{V}\right) k=P1/V=200 kPa5×10−2 cm−3=2000.05 kPa⋅cm3=4000 kPa⋅cm3k = \frac{P}{1/V} = \frac{200 \text{ kPa}}{5 \times 10^{-2} \text{ cm}^{-3}} = \frac{200}{0.05} \text{ kPa} \cdot \text{cm}^3 = 4000 \text{ kPa} \cdot \text{cm}^3

  3. Calculate the pressure using the constant kk.

    P=k(1V)=4000 kPa⋅cm3×0.090009 cm−3P = k \left(\frac{1}{V}\right) = 4000 \text{ kPa} \cdot \text{cm}^3 \times 0.090009 \text{ cm}^{-3} P≈360.036 kPaP \approx 360.036 \text{ kPa}

    The pressure exerted by the gas is approximately 360 kPa\boxed{360 \text{ kPa}}.

Part 5.6

Two properties of an Ideal Gas are:

  • The particles of an ideal gas have negligible volume.
  • There are no intermolecular forces of attraction or repulsion between the particles of an ideal gas.
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