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Difficulty: MediumTemperature Scales and Thermometric Properties

A constant-volume gas thermometer registers a pressure of 50 kPa50\,\text{kPa} at the ice point (0∘C0^\circ\text{C}) and 70 kPa70\,\text{kPa} at the steam point (100∘C100^\circ\text{C}). What is the temperature when the gas pressure measured by the thermometer is 62 kPa62\,\text{kPa}?

  1. A
    12∘C12^\circ\text{C}
  2. 60∘C60^\circ\text{C}Answer
  3. C
    89∘C89^\circ\text{C}
  4. D
    310∘C310^\circ\text{C}

Answer

60∘C60^\circ\text{C}
The temperature θ\theta on the Celsius scale is given by the ratio of the change in thermometric property from the ice point to the total fundamental interval, scaled by 100. Substituting P0=50 kPaP_0 = 50\,\text{kPa}, P100=70 kPaP_{100} = 70\,\text{kPa}, and Pθ=62 kPaP_\theta = 62\,\text{kPa} gives θ=62−5070−50×100=1220×100=60∘C\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}.

Step-by-Step Solution

1
Identify the given thermometric property values at the fixed points and target state
P0=50 kPaP_0 = 50\,\text{kPa}, P100=70 kPaP_{100} = 70\,\text{kPa}, and Pθ=62 kPaP_\theta = 62\,\text{kPa}
These represent the lower fixed point, upper fixed point, and unknown temperature reading respectively.
2
Apply the general thermometric scale conversion formula
θ=Pθ−P0P100−P0×100∘C\theta = \frac{P_\theta - P_0}{P_{100} - P_0} \times 100^\circ\text{C}
Temperature on the Celsius scale varies linearly with the thermometric property relative to fixed points.
3
Substitute the values into the formula and solve for θ\theta
θ=62−5070−50×100=1220×100=60∘C\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}
Carrying out the arithmetic yields the exact temperature of 60∘C60^\circ\text{C}.

Key Concept

Linear interpolation on temperature scales using thermometric properties
Estimated Time:1m 0s
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Topics

Temperature Scales and Thermometric Properties

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