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A learner investigates the relationship between the pressure and volume of an enclosed DIATOMIC gas at 25 °C. He records the volume of the gas for different pressures in the table below. Write down the name of the gas law being investigated.

Physics

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1.1 Write down the name of the gas law being investigated.
The experiment investigates the relationship between pressure and volume of a gas at a constant temperature. This relationship is described by Boyle's Law.
The gas law being investigated is Boyle's Law.

1.2 Two 1V\frac{1}{V} values in the table, (a) and (b), have not been calculated. Calculate these values.

a) For Volume (VV) = 22 cm3^3:

1V=122 cm3\frac{1}{V} = \frac{1}{22 \text{ cm}^3} 1V≈0.045 cm−3\frac{1}{V} \approx \boxed{0.045 \text{ cm}^{-3}}

b) For Volume (VV) = 18 cm3^3:

1V=118 cm3\frac{1}{V} = \frac{1}{18 \text{ cm}^3} 1V≈0.056 cm−3\frac{1}{V} \approx \boxed{0.056 \text{ cm}^{-3}}

1.3 Draw a graph of pressure versus 1V\frac{1}{V} on the attached ANSWER SHEET.
A graph of pressure (PP) versus the inverse of volume (1V\frac{1}{V}) for an ideal gas at constant temperature should be a straight line passing through the origin. This demonstrates the direct proportionality between pressure and 1V\frac{1}{V} as per Boyle's Law.

1.4 Use the graph to determine the volume of the gas at 68 kPa.

  1. Determine the constant of proportionality (kk) from the given data.

    According to Boyle's Law, P=k(1V)P = k \left(\frac{1}{V}\right), where kk is a constant. Using the first two data points from the table (40 kPa, 0.02 cm−3^{-3}) and (80 kPa, 0.04 cm−3^{-3}):

    k=P1V=40 kPa0.02 cm−3=2000 kPa⋅cm3k = \frac{P}{\frac{1}{V}} = \frac{40 \text{ kPa}}{0.02 \text{ cm}^{-3}} = 2000 \text{ kPa} \cdot \text{cm}^3 k=P1V=80 kPa0.04 cm−3=2000 kPa⋅cm3k = \frac{P}{\frac{1}{V}} = \frac{80 \text{ kPa}}{0.04 \text{ cm}^{-3}} = 2000 \text{ kPa} \cdot \text{cm}^3

    The constant kk is 2000 kPa⋅cm32000 \text{ kPa} \cdot \text{cm}^3.

  2. Calculate the volume (VV) at a pressure of 68 kPa.

    P=k(1V)P = k \left(\frac{1}{V}\right) 68 kPa=2000 kPa⋅cm3×1V68 \text{ kPa} = 2000 \text{ kPa} \cdot \text{cm}^3 \times \frac{1}{V} V=2000 kPa⋅cm368 kPaV = \frac{2000 \text{ kPa} \cdot \text{cm}^3}{68 \text{ kPa}} V≈29.41 cm3V \approx \boxed{29.41 \text{ cm}^3}

    1.5.1 Use the conditions at a pressure of 100 kPa and calculate the molar mass of the enclosed gas.
    Given:
    Mass of gas (mm) = 2.49×10−22.49 \times 10^{-2} g
    Pressure (PP) = 100 kPa
    Volume (VV) = 22 cm3^3 (from table at 100 kPa)
    Temperature (TT) = 25 °C
    Ideal gas constant (RR) = 8.314 J⋅mol−1⋅K−18.314 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}

  3. Convert given values to SI units.

    P=100 kPa=100×103 PaP = 100 \text{ kPa} = 100 \times 10^3 \text{ Pa} V=22 cm3=22×(10−2 m)3=22×10−6 m3V = 22 \text{ cm}^3 = 22 \times (10^{-2} \text{ m})^3 = 22 \times 10^{-6} \text{ m}^3 T=25∘C+273.15=298.15 KT = 25^\circ\text{C} + 273.15 = 298.15 \text{ K}
  4. Calculate the number of moles (nn) using the Ideal Gas Law (PV=nRTPV = nRT).

    n=PVRTn = \frac{PV}{RT} n=(100×103 Pa)×(22×10−6 m3)(8.314 J⋅mol−1⋅K−1)×(298.15 K)n = \frac{(100 \times 10^3 \text{ Pa}) \times (22 \times 10^{-6} \text{ m}^3)}{(8.314 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \times (298.15 \text{ K})} n=2.22478.8291n = \frac{2.2}{2478.8291} n≈8.875×10−4 moln \approx 8.875 \times 10^{-4} \text{ mol}
  5. Calculate the molar mass (MM) of the gas.

    M=mnM = \frac{m}{n} M=2.49×10−2 g8.875×10−4 molM = \frac{2.49 \times 10^{-2} \text{ g}}{8.875 \times 10^{-4} \text{ mol}} M≈28.05 g/molM \approx \boxed{28.05 \text{ g/mol}}

    1.5.2 Write down the molecular formula of the enclosed gas.
    The calculated molar mass is approximately 28.05 g/mol28.05 \text{ g/mol}. The problem states the gas is diatomic.
    Comparing this to common diatomic gases:

    • Nitrogen (N2_2) has a molar mass of 2×14.01 g/mol=28.02 g/mol2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol}. The molecular formula of the enclosed gas is N2\boxed{\text{N}_2}.

    1.6 The sketch graph below shows the relationship between volume and temperature for an ideal gas.
    This graph illustrates Charles's Law. It shows that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature (in Kelvin). The linear relationship passing through the origin is characteristic of this law.

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