Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

Exercises · 10.5

Two ideal gas thermometers $A$ and $B$ use oxygen and hydrogen respectively. The following observations are made:

Temperature	Pressure thermometer $A$	Pressure thermometer $B$
Triple-point of water	$1.250 \times 10^{5}\ \text{Pa}$	$0.200 \times 10^{5}\ \text{Pa}$

Normal melting point of sulphur $1.797 \times 10^{5}\ \text{Pa}$ $0.287 \times 10^{5}\ \text{Pa}$ (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers $A$ and $B$ ?

(b) What do you think is the reason behind the slight difference in answers of thermometers $A$ and $B$ ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

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Concept understanding — Temperature Scale Conversion

Temperature Scale Conversion

You already know temperature as a measure of hotness or coldness. But here's the thing: different countries and different scientific fields measure that same hotness using different numbers. Water boils at 100 on the Celsius scale, but at 212 on the Fahrenheit scale. That's not because the water is different — it's because the rulers are different.

The core idea

A temperature scale is just a number line someone decided to draw on the same physical reality. Converting between scales means finding the number on one number line that corresponds to the same physical hotness as a given number on another number line.

Think of it like this: if you measure a table's length in feet and get 6, and I measure it in metres and get 1.83, we're both describing the same table. Temperature conversion works exactly the same way — same physical state, different numbers.

The three main scales you'll meet

Scale	Freezing point of water	Boiling point of water	Used by
Celsius (°C)	0	100	Most of the world, science
Fahrenheit (°F)	32	212	USA, a few other countries
Kelvin (K)	273.15	373.15	All scientific work

Notice something: Celsius and Kelvin have the same step size — a change of 1°C is exactly a change of 1 K. They just start at different places. Fahrenheit has a smaller step size (180 steps between freezing and boiling, instead of 100).

The conversion formulas

°F = \frac{9}{5}(°C) + 32

These aren't magic. Each one comes from the simple idea of matching two number lines.

Why the formulas look like that

Between freezing and boiling water:

Celsius has 100 steps (0 to 100)
Fahrenheit has 180 steps (32 to 212)

So one Celsius step is $\frac{180}{100} = \frac{9}{5}$ Fahrenheit steps. That's the $\frac{9}{5}$ factor. The $+32$ just shifts the starting point — because 0°C doesn't correspond to 0°F, it corresponds to 32°F.

For Kelvin: since the step size is identical to Celsius, you just add the offset. The 273.15 comes from the fact that absolute zero (the coldest possible temperature) is 0 K, which is $-273.15$ °C.

A worked example

Convert 25°C to Fahrenheit.

Step 1: Multiply by $\frac{9}{5}$ .

25 \times \frac{9}{5} = 25 \times 1.8 = 45

Step 2: Add 32.

45 + 32 = 77

So 25°C = 77°F. A warm spring day in Celsius terms is 77°F — same weather, different number.

Tip

For quick mental estimates: double the Celsius and add 30. It's not exact (you get 80 instead of 77 here), but it's close enough for everyday "should I wear a jacket?" decisions.

The one thing students mess up …

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Using the ideal gas temperature scale $T = 273.16 \times (P/P_{tp})$ , thermometer $A$ gives $T_A \approx 392.69\ \text{K}$ and thermometer $B$ gives $T_B \approx 391.94\ \text{K}$ . The slight difference arises because real gases deviate from ideal behaviour; extrapolating to zero pressure eliminates the discrepancy.

Why this approach works

The ideal gas thermometer defines temperature by the relation $T = 273.16 \times \frac{P}{P_{tp}}$ , where $P_{tp}$ is the pressure at the triple point of water (273.16 K). This works because for an ideal gas, pressure is directly proportional to absolute temperature at constant volume. Real gases like oxygen and hydrogen approximate this, but only at low pressures — at the pressures used here, small non-ideal effects cause slight differences.

The problem gives us two gases at two different pressures, so we can compute the sulphur melting point for each thermometer separately. The slight mismatch between $T_A$ and $T_B$ is not a fault; it’s a real physical effect due to intermolecular forces and finite molecular size.

Step-by-step calculation

1. Write the ideal gas temperature formula

For a constant-volume gas thermometer:

T = 273.16 \times \frac{P}{P_{tp}}

where $P_{tp}$ is the pressure at the triple point of water (273.16 K).

2. Compute for thermometer $A$ (oxygen)

Triple-point pressure: $P_{tp,A} = 1.250 \times 10^{5}\ \text{Pa}$

Pressure at sulphur melting point: $P_A = 1.797 \times 10^{5}\ \text{Pa}$

T_A = 273.16 \times \frac{1.797 \times 10^{5}}{1.250 \times 10^{5}} = 273.16 \times 1.4376

T_A = 392.69\ \text{K}

3. Compute for thermometer $B$ (hydrogen)

Triple-point pressure: $P_{tp,B} = 0.200 \times 10^{5}\ \text{Pa}$

Pressure at sulphur melting point: $P_B = 0.287 \times 10^{5}\ \text{Pa}$

T_B = 273.16 \times \frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} = 273.16 \times 1.435

T_B = 391.94\ \text{K}

Watch out

A common mistake is to forget that the triple-point pressure is different for each thermometer — they are separate instruments with different gas fillings. Using the wrong $P_{tp}$ will give a wrong temperature.

4. Compare the two results

Thermometer $A$ reads $392.69\ \text{K}$ , thermometer $B$ reads $391.94\ \text{K}$ . The difference is about $0.75\ \text{K}$ — small but noticeable.

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Why this approach works

Step-by-step calculation

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