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    The apparatus depicted in the diagram below was used by a student to examine how the pressure

    Question
    SLPaper 2

    The apparatus depicted in the diagram below was used by a student to examine how the pressure ppp of air varies with volume while maintaining a constant temperature.

    The air was confined in a tube with a uniform cross-sectional area, positioned above a column of oil.

    Image

    The pump pushes oil up the tube, reducing the volume of the trapped air.

    1.

    The student measured the height HHH of the air column and the corresponding air pressure ppp. After each decrease in volume, they paused for a while before recording the pressure. Outline why this step was necessary.

    [1]
    Verified
    Solution

    in order to keep the temperature constant 1 mark

    in to allow the system to reach thermal equilibrium with the surroundings 1 mark

    2.

    The following graph of ppp versus 1H\frac{1}{H}H1​ was obtained. Error bars were negligibly small.

    Image

    The equation of the line of best fit is p=a+bHp = a + \frac{b}{H}p=a+Hb​.

    Determine the value of bbb including an appropriate unit.

    [3]
    Verified
    Solution

    recognizes bbb as gradient 1 mark

    calculates bbb in range 5.8×1045.8 \times 10^{4}5.8×104 to 6.2×1046.2 \times 10^{4}6.2×104 1 mark

    Unit: Pa m\text{Pa m}Pa m 1 mark

    3.

    Outline how the results of this experiment are consistent with the ideal gas law at constant temperature.

    [2]
    Verified
    Solution

    V∝HV \propto HV∝H thus ideal gas law gives p∝1Hp \propto \frac{1}{H} p∝H1​ 1 mark

    so graph should be a straight line through origin, as observed 1 mark

    4.

    The cross-sectional area of the tube is 1.15×10−31.15 × 10^{-3}1.15×10−3 m2\mathrm{m}^2m2 and the temperature of air is 295 K. Estimate the number of moles of air in the tube.

    [2]
    Verified
    Solution

    n=bARTn=\frac{b A}{R T}n=RTbA​ OR correct substitution of one point from the graph 1 mark

    n=6×104×1.15×10−38.31×295=0.028≈0.03n=\frac{6 \times 10^{4} \times 1.15 \times 10^{-3}}{8.31 \times 295}=0.028 \approx 0.03n=8.31×2956×104×1.15×10−3​=0.028≈0.03 1 mark

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    5.

    The equation in may be used to predict the pressure of the air at extremely large values of 1H\frac{1}{H}H1​. Suggest why this will be an unreliable estimate of the pressure.

    [2]
    Verified
    Solution

    very large 1H\frac{1}{H}H1​ means very small volumes / very high pressures 1 mark

    at very small volumes the ideal gas does not apply

    OR

    at very small volumes some of the assumptions of the kinetic theory of gases do not hold 1 mark

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