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I) The gas law under investigation is Gay-Lussac's Law (also known as the Pressure Law).

Physics

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Answer

b

127956 Pa\text{127956 Pa}

Step-by-step solution

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Part a(i)

The gas law under investigation is Gay-Lussac's Law (also known as the Pressure Law).

ii) To ensure that all the air trapped in the bulb was at the temperature recorded by the thermometer, the thermometer should be placed inside the water bath, close to the bulb. Additionally, the water in the bath should be stirred continuously to ensure uniform temperature distribution throughout the water and the air bulb.

Part b

127956 Pa\text{127956 Pa}

Given:
Height difference in mercury h=20 cm=0.20 mh = 20 \text{ cm} = 0.20 \text{ m}
Density of mercury ρ=13600 kg/m3\rho = 13600 \text{ kg/m}^3
Acceleration due to gravity g=9.8 m/s2g = 9.8 \text{ m/s}^2
Assume standard atmospheric pressure Patm=1.013×105 PaP_{atm} = 1.013 \times 10^5 \text{ Pa}.

  1. Calculate the pressure exerted by the mercury column.

    The pressure difference ΔP\Delta P due to the mercury column is given by the formula ΔP=ρgh\Delta P = \rho g h.

    ΔP=(13600 kg/m3)×(9.8 m/s2)×(0.20 m)\Delta P = (13600 \text{ kg/m}^3) \times (9.8 \text{ m/s}^2) \times (0.20 \text{ m}) ΔP=26656 Pa\Delta P = 26656 \text{ Pa}
  2. Determine the total pressure of the air trapped in the bulb.

    From the diagram, the mercury level in the arm connected to the bulb is lower than the level in the arm open to the atmosphere. This indicates that the pressure of the air in the bulb is greater than the atmospheric pressure.

    Pbulb=Patm+ΔPP_{bulb} = P_{atm} + \Delta P Pbulb=1.013×105 Pa+26656 PaP_{bulb} = 1.013 \times 10^5 \text{ Pa} + 26656 \text{ Pa} Pbulb=101300 Pa+26656 PaP_{bulb} = 101300 \text{ Pa} + 26656 \text{ Pa} Pbulb=127956 PaP_{bulb} = 127956 \text{ Pa}

    The total pressure of air trapped in the bulb is 127956 Pa\boxed{\text{127956 Pa}}.

Part c

The graph of pressure versus temperature (in ∘C^\circ\text{C}) for a fixed mass of gas at constant volume (Gay-Lussac's Law) is a straight line with a positive slope. It does not pass through the origin (0,0)(0,0) because 0∘C0^\circ\text{C} is not absolute zero. If extrapolated, the line would intersect the temperature axis at approximately −273.15∘C-273.15^\circ\text{C}.

\begin{tikzpicture} \draw[->] (0,0) -- (5,0) node[below] {Temperature ($^\circ$C)}; \draw[->] (0,0) -- (0,4) node[left] {Pressure ($\text{N/m}^2$)}; \draw[thick, blue] (1,1) -- (4,3.5); % Example line segment \end{tikzpicture}

(Note: The graph should be a straight line with a positive slope, starting from a point in the positive temperature range and extending upwards. It should not pass through the origin.)

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