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Gay-Lussac: why Kelvin matters

Observe a real run and compare what happens when P/T uses temperature in Celsius and in kelvin.

  • Gay-Lussac's Law
  • 38 min
  • secondary school
  • English
  • Physics · Chemistry
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Visual scale relating Celsius and kelvin with the formula temperature in kelvin equals temperature in Celsius plus 273.15.
Gay-Lussac's Law

Learning Outcomes

  • Convert Celsius temperatures to kelvin in real lab data.

  • Calculate and compare several P/T(K) values.

  • Explain why the gas-law ratio uses absolute temperature.

More for this lab

Student activity preview

Activity Content

Preview only. In a class session, students can fill in responses and submit their work to the teacher.

1

The scale changes the meaning

5 min

Is gas at 40 °C “twice as hot” as gas at 20 °C? No. The Celsius scale places 0 °C at a point related to water; moving that zero would not change the physical state of the gas. In fact, 20 °C and 293.15 K describe exactly the same temperature.

This difference matters when we use a ratio. In Gay-Lussac's law, we compare pressure and temperature through P/T. For that ratio to have physical meaning, temperature must be measured from a zero that is not a conventional choice. The kelvin scale starts at the lower limit of temperature, 0 K. It therefore lets us compare absolute temperatures, whereas “twice as many degrees Celsius” does not mean twice the absolute temperature.

Two teams will now analyze the same heated container. One will calculate the ratio using Celsius and the other using kelvin. You will check how much each result changes and use lab data to decide which scale better represents the pressure-temperature relationship.

From Celsius to kelvin

Visual scale relating Celsius and kelvin with the formula temperature in kelvin equals temperature in Celsius plus 273.15.

Use the figure as a conversion reminder. Each pair of marks represents the same temperature on the two scales. In the table, use K = °C + 273.15.

The ideal model would treat P/T(K) as constant. What do you predict for real readings?

Justify your choice in one sentence. You may refer to the ideal model or uncertainty in real measurements.

2

Observe where the data come from

12 min

Open the lab and observe the heating of the same fixed-volume ethanol sample. Before starting, prepare two rows on paper or in a note with columns for time, temperature, and pressure. Near 00:30 and 03:30, record temperature in °C and pressure in kPa, then convert both temperatures to kelvin. The activity also includes rows from the same run for comparison.

Open the Gay-Lussac lab

  1. The sealed sample always contains the same total amount of ethanol, and the container keeps a fixed volume. The ideal model also treats the gas-phase amount as fixed; the lab records temperature and pressure, not the amount in each phase.

  2. Open the lab from this activity's lab button.

  3. On the configuration screen, select the only available sample: 0.014 mol of ethanol.

  4. Start one observation, which takes about 4 minutes. During the run, record approximate temperature and pressure at the times shown in the following table.

  5. Return to TEACH and complete that table before continuing with the analysis.

Lab readings

The times are already provided. Copy observed readings into the °C and kPa columns. Units appear in the headings. Then convert each temperature to K and round to one decimal place. Finally, calculate each P/T(K) and keep four decimal places.

Run time (min:s) Temperature °C Pressure kPa Temperature K P/T(K) kPa/K

After completing the table, what do you observe when comparing your two P/T(K) values?

3

Compare the ratios

13 min

The list contains three pre-calculated reference points from the same run. Celsius temperatures were rounded to one decimal first, then 273.15 was added and kelvin was rounded to one decimal; P/T is shown to four decimals. Use it for reference. Your own two ratios are still the ones you calculated from the lab readings.

Reference data
- Measurement 1: T = 23.9 °C = 297.1 K; P = 87.76 kPa; P/T = 0.2954 kPa/K.
- Measurement 41: T = 34.9 °C = 308.1 K; P = 91.36 kPa; P/T = 0.2965 kPa/K.
- Measurement 61: T = 44.6 °C = 317.8 K; P = 95.48 kPa; P/T = 0.3004 kPa/K.

Celsius calculation. Complete this format: Initial ratio = ___ kPa/°C; final ratio = ___ kPa/°C; relative change = ___ %. Use 87.76 / 23.9, 95.48 / 44.6, and |last - first| / first × 100. Round ratios to two decimals and the percentage to one decimal place.

Kelvin change. Use the first and last ratios in the reference list to calculate |last - first| / first × 100. Show the operation and percentage, rounded to one decimal place.

Which temperature scale produces the more stable calculated ratio?

Justify the previous choice by comparing the relative change you calculated with Celsius and the one you calculated with kelvin from the reference list. Do not calculate them again.

4

Decide which scale better represents the relationship

8 min

You have calculated ratios using two scales that represent the same temperatures but place zero at different points. Use ratio stability and the position of each zero to build your conclusion. Limit the response to comparing temperature scales; small differences between real readings mean that you should not claim an exact constant.

Conclusion. In 2-3 sentences, state which scale is appropriate for the law and explain why its physical zero matters. Use at least one percentage as evidence.

Compare the result of your two P/T(K) ratios with your initial prediction. What do you do with the prediction?

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