Measuring Temperature — Question 1

Section 18-3 · Physics pages 2852–2853

Question 1 Absolute pressure matters in a gas thermometer A sealed constant-volume ideal gas thermometer obeys p/T = constant with absolute pressure. At a reference temperature 300 K, its gauge pressure is 20.0 kPa above ambient 100.0 kPa. In an unknown bath, the gauge reads 44.0 kPa with unchanged ambient pressure. sealed gas pressure gauge Schematic; labels, not drawing lengths, specify the data. (a) Find the bath temperature. Quantify the result from incorrectly taking the ratio of gauge pressures. (b) If ambient pressure in the second measurement is instead 98.0 kPa with the same gauge reading, infer the corrected bath temperature.
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Question 1 - Solution Absolute pressure matters in a gas thermometer sealed gas pressure gauge (a) Method and calculation pr = 120 kPa, p = 144 kPa, T = 300144 120 = 360 K. Tgauge = 30044 20 = 660 K. Gauge zero is an atmospheric reference, not zero molecular pressure. (b) Interpretation and further calculation p = 98 + 44 = 142 kPa, T = 300142 120 = 355 K. Ignoring the ambient change produces Tuncorrected −T = 360 −355 = 5 K. Physical check Ratios of absolute pressures remove the gas amount and fixed volume; ratios of shifted gauge readings do not.

Question 1

Absolute pressure matters in a gas thermometer

A sealed constant-volume ideal gas thermometer obeys p/T=constantp/T=\text{constant} with absolute pressure. At a reference temperature 300K300\,\mathrm K, its gauge pressure is 20.0kPa20.0\,\mathrm{kPa} above ambient 100.0kPa100.0\,\mathrm{kPa}. In an unknown bath, the gauge reads 44.0kPa44.0\,\mathrm{kPa} with unchanged ambient pressure.

Diagram shown in the original SVG worksheet.

  1. Find the bath temperature. Quantify the result from incorrectly taking the ratio of gauge pressures.

  2. If ambient pressure in the second measurement is instead 98.0kPa98.0\,\mathrm{kPa} with the same gauge reading, infer the corrected bath temperature.

Question 1 - Solution

Absolute pressure matters in a gas thermometer

Diagram shown in the original SVG worksheet.

(a) Method and calculation

pr=120kPa,p=144kPa,T=300144120=360K.p_r=120\,\mathrm{kPa},\quad p=144\,\mathrm{kPa},\quad T=300\frac{144}{120}=360\,\mathrm K.Tgauge=3004420=660K.T_{\mathrm{gauge}}=300\frac{44}{20}=660\,\mathrm K.Gauge zero is an atmospheric reference, not zero molecular pressure.

(b) Interpretation and further calculation

p=98+44=142kPa,T=300142120=355K.p=98+44=142\,\mathrm{kPa},\qquad T=300\frac{142}{120}=355\,\mathrm K.Ignoring the ambient change producesTuncorrected−T=360−355=5K.T_{\mathrm{uncorrected}}-T=360-355=5\,\mathrm K.

Physical check

Ratios of absolute pressures remove the gas amount and fixed volume; ratios of shifted gauge readings do not.