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Question: A barometer reads 75.0 cm on a steel scale. The room temperature is 30°C. The scale is correctly gra...

A barometer reads 75.0 cm on a steel scale. The room temperature is 30°C. The scale is correctly graduated for 0°C. The coefficient of linear expansion of steel is α=1.2×10−5∘C−1\alpha = 1.2 \times 10^{-5} {^\circ}C^{-1} and the coefficient of volume expansion of mercury is γ=1.8×10−4∘C−1\gamma = 1.8 \times 10^{-4} {^\circ}C^{-1}. Find the correct atmospheric pressure.

A

75.0 cmHg

B

74.6 cmHg

C

75.4 cmHg

D

75.1 cmHg

Answer

74.6 cmHg

Explanation

Solution

The problem requires correcting a barometer reading for the thermal expansion of both the measuring scale and the fluid (mercury).

  1. Thermal Expansion of the Steel Scale: The steel scale is calibrated at 0°C. At a higher temperature (30°C), the scale expands. A reading of 75.0 cm on the expanded scale means the physical length of the mercury column is greater than 75.0 cm. If hobsh_{obs} is the observed reading and ΔT\Delta T is the temperature difference from calibration, the actual physical height of the mercury column (hmerc,30h_{merc, 30}) at 30°C is given by: hmerc,30=hobs(1+αΔT)h_{merc, 30} = h_{obs} (1 + \alpha \Delta T) where α\alpha is the coefficient of linear expansion of steel.

  2. Thermal Expansion of Mercury: Mercury also expands with temperature. The height of the mercury column at 30°C (hmerc,30h_{merc, 30}) is related to its height at 0°C (htrue,0h_{true, 0}) by: hmerc,30=htrue,0(1+γΔT)h_{merc, 30} = h_{true, 0} (1 + \gamma \Delta T) where γ\gamma is the coefficient of volume expansion of mercury. The true atmospheric pressure is represented by the height of the mercury column at the calibration temperature (0°C).

  3. Combining the Effects: Equating the two expressions for hmerc,30h_{merc, 30}: hobs(1+αΔT)=htrue,0(1+γΔT)h_{obs} (1 + \alpha \Delta T) = h_{true, 0} (1 + \gamma \Delta T)

    Solving for the true atmospheric pressure (htrue,0h_{true, 0}): htrue,0=hobs1+αΔT1+γΔTh_{true, 0} = h_{obs} \frac{1 + \alpha \Delta T}{1 + \gamma \Delta T}

  4. Calculation: Given: hobs=75.0h_{obs} = 75.0 cm ΔT=30∘C−0∘C=30∘C\Delta T = 30^\circ\text{C} - 0^\circ\text{C} = 30^\circ\text{C} α=1.2×10−5 ∘C−1\alpha = 1.2 \times 10^{-5} \, {^\circ}\text{C}^{-1} γ=1.8×10−4 ∘C−1\gamma = 1.8 \times 10^{-4} \, {^\circ}\text{C}^{-1}

    Calculate the expansion terms: αΔT=(1.2×10−5)×30=3.6×10−4=0.00036\alpha \Delta T = (1.2 \times 10^{-5}) \times 30 = 3.6 \times 10^{-4} = 0.00036 γΔT=(1.8×10−4)×30=5.4×10−3=0.0054\gamma \Delta T = (1.8 \times 10^{-4}) \times 30 = 5.4 \times 10^{-3} = 0.0054

    Substitute these values into the formula: htrue,0=75.0 cm×1+0.000361+0.0054h_{true, 0} = 75.0 \, \text{cm} \times \frac{1 + 0.00036}{1 + 0.0054} htrue,0=75.0 cm×1.000361.0054h_{true, 0} = 75.0 \, \text{cm} \times \frac{1.00036}{1.0054} htrue,0=75.0 cm×0.99503680569...h_{true, 0} = 75.0 \, \text{cm} \times 0.99503680569... htrue,0≈74.62776 cmh_{true, 0} \approx 74.62776 \, \text{cm}

Rounding to three significant figures (consistent with the input 75.0 cm), the correct atmospheric pressure is 74.6 cmHg.