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Measurement and Data Processing - Graphical techniques

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The independent variable is plotted on the xx-axis, while the dependent variable is plotted on the yy-axis.

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A line of best fit (trend line) should be drawn to represent the mathematical relationship, which may be linear or a curve. It does not necessarily pass through all data points but should minimize the distance from them.

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The gradient (mm) of a linear graph is calculated using the change in yy divided by the change in xx, often represented as ΔyΔx\frac{\Delta y}{\Delta x}.

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Interpolation is the process of estimating a value within the range of measured data points, whereas extrapolation involves extending the line of best fit beyond the experimental data range.

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Error bars are used to represent the absolute uncertainty of the measurements on a graph. A line of best fit should ideally pass through all error bars.

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To calculate the uncertainty in the gradient, lines of maximum and minimum slope (lines of worst fit) are drawn. These must still pass through all error bars.

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Direct proportionality is indicated by a straight line passing through the origin (0,0)(0,0), satisfying the equation y=kxy = kx.

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Systematic errors can often be identified by a non-zero intercept on the yy-axis when the theoretical relationship predicts an intercept of zero.

📐Formulae

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

y=mx+cy = mx + c

Uncertainty in gradient=mmax−mmin2\text{Uncertainty in gradient} = \frac{m_{max} - m_{min}}{2}

Uncertainty in intercept=cmax−cmin2\text{Uncertainty in intercept} = \frac{c_{max} - c_{min}}{2}

PV=nRT  ⟹  P=nRT(1V)PV = nRT \implies P = nRT \left(\frac{1}{V}\right)

💡Examples

Problem 1:

A student measures the volume VV of a gas at different temperatures TT in Kelvin, keeping pressure PP constant. The resulting graph of VV against TT is a straight line. If the gradient of the graph is 0.082 dm3K−10.082 \text{ dm}^3\text{K}^{-1} and the pressure is 101.3 kPa101.3 \text{ kPa}, determine the number of moles nn of the gas. (Use R=8.31 J K−1mol−1R = 8.31 \text{ J K}^{-1}\text{mol}^{-1} and note 1 dm3=10−3 m31 \text{ dm}^3 = 10^{-3} \text{ m}^3).

Solution:

  1. From the Ideal Gas Law: V=(nRP)TV = \left(\frac{nR}{P}\right)T.
  2. The gradient m=nRPm = \frac{nR}{P}.
  3. Convert pressure to Pascals: P=101.3×103 PaP = 101.3 \times 10^3 \text{ Pa}.
  4. Convert gradient to m3K−1\text{m}^3\text{K}^{-1}: m=0.082×10−3 m3K−1m = 0.082 \times 10^{-3} \text{ m}^3\text{K}^{-1}.
  5. Rearrange for nn: n=m⋅PR=(0.082×10−3)×(101.3×103)8.31n = \frac{m \cdot P}{R} = \frac{(0.082 \times 10^{-3}) \times (101.3 \times 10^3)}{8.31}.
  6. n=8.30668.31≈1.00 moln = \frac{8.3066}{8.31} \approx 1.00 \text{ mol}.

Explanation:

The gradient of a VV vs TT graph represents nRP\frac{nR}{P} according to Charles's Law and the Ideal Gas Equation. By substituting the known values into the gradient expression, the amount of substance can be calculated.

Problem 2:

A student determines the density of a liquid by plotting mass (mm) on the yy-axis against volume (VV) on the xx-axis. The best-fit line gradient is 0.85 g cm−30.85 \text{ g cm}^{-3}. The maximum possible gradient is 0.88 g cm−30.88 \text{ g cm}^{-3} and the minimum is 0.82 g cm−30.82 \text{ g cm}^{-3}. Calculate the density with its absolute uncertainty.

Solution:

  1. Density ρ=gradient=0.85 g cm−3\rho = \text{gradient} = 0.85 \text{ g cm}^{-3}.
  2. Absolute uncertainty Δρ=mmax−mmin2=0.88−0.822=0.03 g cm−3\Delta \rho = \frac{m_{max} - m_{min}}{2} = \frac{0.88 - 0.82}{2} = 0.03 \text{ g cm}^{-3}.
  3. Final value: 0.85±0.03 g cm−30.85 \pm 0.03 \text{ g cm}^{-3}.

Explanation:

The uncertainty in the gradient of a graph is found by taking half the difference between the steepest and shallowest possible lines of best fit that pass through the error bars.

Graphical techniques - Revision Notes & Key Formulas | IB Grade 11 Chemistry