Why Gas-Law Temperatures Must Be Measured in Kelvin

A sealed bottle of gas is cooled to \(-20.0^\circ\text{C}\), then warmed to \(40.0^\circ\text{C}\). A student tries to calculate the pressure change using the ratio \(40/-20\), obtaining \(-2\).

Something has gone seriously wrong. The student has predicted a negative pressure simply because the temperature crossed zero on the Celsius scale.

The problem isn’t the algebra. It’s the temperature scale.

Gas-law temperatures must be measured in kelvin because gas-law relationships depend on absolute temperature, measured from absolute zero. The zero on the Celsius scale is an arbitrary reference point based on water freezing. It has nothing to do with the minimum possible thermodynamic temperature.

This is why every temperature used in the ideal gas equation, and in gas-law ratios, must be converted to kelvin before you calculate.

01Why Celsius gives the wrong answer

Think about what happens when you heat a gas.

Its particles move faster, transferring momentum to the container walls through more frequent and more energetic collisions. Depending on the experimental conditions, the gas either expands or develops a higher pressure.

The important relationship is between the gas’s behaviour and its absolute temperature.

Consider the difference between these two temperature scales:

• Celsius sets \(0^\circ\text{C}\) at the freezing point of water.
• Kelvin sets \(0\text{ K}\) at absolute zero, the lowest possible thermodynamic temperature.

The two scales have equally sized intervals. A temperature increase of \(10^\circ\text{C}\) is exactly the same temperature increase as \(10\text{ K}\).

However, their zero points are different.

The conversion is:

\[ T(\text{K})=t(^{\circ}\text{C})+273.15 \]

where:

• \(T\) is absolute temperature in kelvin.
• \(t\) is temperature in degrees Celsius.

For example:

The problem with using Celsius in a ratio

Imagine heating a gas from \(15^\circ\text{C}\) to \(45^\circ\text{C}\), while keeping its pressure and amount constant.

A student might notice that the Celsius temperature has tripled.

They might therefore predict that the gas volume triples.

But this is incorrect.

Convert both temperatures:

\[ T_1=15+273.15=288.15\text{ K} \]

\[ T_2=45+273.15=318.15\text{ K} \]

The actual temperature ratio is:

\[ \frac{T_2}{T_1} = \frac{318.15}{288.15} =1.104 \]

The gas volume increases by approximately 10.4%, not 200%.

Why?

Because Celsius measures temperature relative to an arbitrary starting point. Doubling or tripling a number on an offset scale does not tell you how much the underlying physical quantity has changed.

Think of a ruler whose zero mark has been shifted. The distance between two marks stays the same, but ratios calculated from their positions become misleading.

The analogy has one limitation: temperature is not literally a distance. The useful idea is that differences survive a shift in the zero point, but ratios do not.

What about temperature changes?

There is an important distinction between a temperature and a temperature change.

Because both scales have the same interval size:

\[ \Delta T(\text{K})=\Delta t(^{\circ}\text{C}) \]

For example, heating a gas from \(20^\circ\text{C}\) to \(60^\circ\text{C}\) gives:

\[ \Delta T=40\text{ K} \]

You do not need to add 273.15 when calculating a temperature difference.

However, if you are substituting temperatures into a gas-law equation or calculating a temperature ratio, you must use kelvin.

This distinction is a common source of lost marks.

02How scientists discovered absolute zero

The reason we use kelvin is not just a mathematical convention. It comes from experimental observations of how gases behave when their temperature changes.

Scientists wanted to answer an important question:

Could there be a lowest possible temperature?

During the eighteenth and nineteenth centuries, researchers investigated how gas pressure and volume changed with temperature. Their measurements suggested that gas properties approached zero at approximately \(-273^\circ\text{C}\) when the experimental trends were extrapolated.

Researchers including Guillaume Amontons and Joseph-Louis Gay-Lussac contributed to this work. In 1848, William Thomson, later known as Lord Kelvin, published a paper proposing an absolute thermometric scale.

The experimental reasoning is easier to understand if we follow a gas sample through a temperature investigation.

Experiment 1: Measuring gas volume at constant pressure

Imagine a gas trapped inside a graduated syringe.

The syringe has a movable, low-friction plunger. Its tip is sealed, so no gas can enter or escape. The syringe barrel sits inside a water bath, alongside a thermometer.

The plunger is exposed to atmospheric pressure, so it can move freely as the gas expands or contracts. We assume its friction and weight are negligible.

The amount of gas remains constant, and the gas pressure stays approximately constant.

Now increase the temperature of the water bath.

What should happen to the gas?

Its particles move faster as its temperature increases. The gas expands, pushing the plunger outwards.

The experimenter records the gas volume at several temperatures after allowing the system to reach thermal equilibrium.

The following is an illustrative dataset, not a historical experimental record.

The measurements show an approximately linear increase in volume as Celsius temperature increases.

What does the experiment predict?

If we plot volume against Celsius temperature, the data form an approximately straight line.

We can extend that line backwards to estimate the temperature at which the gas volume would theoretically reach zero.

Let’s estimate the intercept using two measurements.

At \(0^\circ\text{C}\), the volume is \(100.0\text{ mL}\).

At \(50^\circ\text{C}\), the volume is \(118.3\text{ mL}\).

The gradient is:

\[ \begin{aligned} m&=\frac{118.3-100.0}{50-0}\\ &=0.366\text{ mL}/^\circ\text{C} \end{aligned} \]

Using the linear relationship:

\[ V=100.0+0.366t \]

where \(V\) is volume in millilitres and \(t\) is temperature in degrees Celsius.

Set the volume equal to zero:

\[ \begin{aligned} 0&=100.0+0.366t\\ t&=\frac{-100.0}{0.366}\\ t&\approx-273.2^\circ\text{C} \end{aligned} \]

The extrapolated temperature is approximately \(-273^\circ\text{C}\).

This is remarkably close to the modern value of absolute zero:

\[ 0\text{ K}=-273.15^\circ\text{C} \]

Why wasn’t absolute zero measured directly?

There is a catch.

The experiment does not actually show a gas shrinking to zero volume.

As a real gas is cooled, it will generally condense into a liquid before reaching absolute zero. Once condensation occurs, the simple gas-volume relationship no longer describes the entire sample.

The zero-volume result is therefore an extrapolation: we extend the approximately linear behaviour of a gas beyond the range in which that gas remains gaseous.

This is an important scientific distinction.

The experiment provides evidence for a common absolute temperature scale. It does not demonstrate that a real gas can physically reach zero volume.

Experiment 2: Improving the gas thermometer

The constant-pressure syringe experiment is useful, but it has limitations.

Real gases do not behave perfectly ideally. Their particles occupy space and experience intermolecular attractions.

These effects can influence the experimental relationship between temperature and volume.

A complementary design uses a constant-volume gas thermometer.

In this arrangement:

• A bulb contains a fixed amount of gas.
• The bulb is placed inside a temperature-controlled bath.
• A pressure sensor or manometer measures the gas pressure.
• The gas volume is kept constant.
• The experimenter changes the bath temperature and records the corresponding pressure.

At constant volume and fixed gas amount, pressure increases as temperature increases.

A graph of pressure against Celsius temperature also produces an approximately straight line that extrapolates towards zero pressure near \(-273.15^\circ\text{C}\).

However, scientists can improve the measurement further.

They repeat the experiment using lower gas pressures. At lower pressures, particles are further apart, so their intermolecular interactions and individual volumes have less influence on the measured behaviour.

The results approach ideal-gas behaviour more closely.

This process of extrapolating measurements towards zero gas pressure helps reduce the influence of real-gas effects.

It also explains why the results from different gases can converge towards the same absolute temperature.

The important distinction is that Kelvin is not defined by this experiment today. Since 2019, the SI definition of the kelvin has been based on fixing the Boltzmann constant at an exact value, connecting thermodynamic temperature to energy.

You can read about this modern definition through the Bureau International des Poids et Mesures (BIPM), and the historical development through NIST’s history of the kelvin.

03Connecting Kelvin to the ideal gas equation

Now the mathematical reason for using kelvin should be clearer.

The ideal gas equation is:

\[ PV=nRT \]

where:

The temperature must be in kelvin.

The reason becomes clear when we rearrange the equation.

\[ P=\frac{nRT}{V} \]

If the amount of gas and volume remain constant, all the other quantities are constant except pressure and temperature.

Therefore:

\[ P\propto T \]

This relationship only works when \(T\) is measured from absolute zero.

Similarly, rearranging for volume gives:

\[ V=\frac{nRT}{P} \]

At constant pressure and fixed gas amount:

\[ V\propto T \]

Again, temperature must be measured in kelvin.

Using Celsius would shift the zero point away from the origin of these proportional relationships.

Choosing consistent units

Kelvin is required regardless of which value of \(R\) you use.

However, the units of pressure and volume must match the units in your gas constant.

For the common HSC Chemistry combination:

\[ R=8.314\text{ kPa L mol}^{-1}\text{K}^{-1} \]

Use:

• Pressure in kPa.
• Volume in L.
• Amount in mol.
• Temperature in K.

Alternatively, if pressure is measured in pascals and volume in cubic metres:

\[ R=8.314\text{ J mol}^{-1}\text{K}^{-1} \]

This works because:

\[ 1\text{ Pa m}^3=1\text{ J} \]

Notice that kelvin appears in the units of both versions of \(R\).

You cannot substitute degrees Celsius and expect the units to work correctly.

One more detail: gas-law pressure must be absolute pressure, measured relative to a perfect vacuum, rather than gauge pressure measured relative to the surrounding atmosphere.

The conversion is:

\[ P_{\text{absolute}} = P_{\text{gauge}}+P_{\text{atmospheric}} \]

For example, a pressure gauge reading of 150 kPa at an atmospheric pressure of 101 kPa corresponds to an absolute pressure of 251 kPa.

This is a separate issue from temperature conversion, but both mistakes can produce incorrect gas-law calculations.

For more detail on choosing the correct gas constant, see Choosing R When Pressure Is in kPa and Volume Is in L.

04Ideal gas assumptions versus real-gas corrections

Using kelvin does not automatically make every gas behave ideally.

The ideal gas equation is a mathematical model. It assumes that:

• Gas particles have negligible volume compared with the container.
• There are no significant intermolecular attractions or repulsions between particles.
• Collisions between particles and the container walls are elastic.

Real gases do not satisfy these assumptions perfectly.

The ideal gas model is generally more accurate under conditions of:

• Low pressure, where particles are relatively far apart.
• High temperature, where intermolecular attractions have less influence on the gas behaviour.

At high pressures, the actual volume occupied by gas molecules becomes more significant.

At low temperatures, intermolecular attractions become more influential, and gases may condense.

These effects cause real gases to deviate from ideal gas predictions.

Is converting to kelvin an experimental correction?

No. This is a distinction worth remembering.

Kelvin is required even for a perfectly ideal gas.

A real-gas correction, on the other hand, accounts for the fact that actual particles occupy space and interact with one another.

For example, the van der Waals equation introduces corrections for molecular volume and intermolecular attractions.

These corrections improve the model under conditions where the ideal gas equation becomes less accurate.

They do not replace the need for kelvin.

In experimental work, scientists can also reduce deviations by using lower gas pressures or extrapolating measurements towards zero pressure.

For HSC calculations, if a question specifies an ideal gas, use \(PV=nRT\). If it asks you to discuss experimental limitations, explain the assumptions and why real gases may deviate. Do not invent an additional correction unless the question supplies an appropriate model.

For a broader explanation of these assumptions and the relationships between gas variables, see Ideal Gas Law and Gas Relationships for HSC Chemistry.

05Worked examples

06Questions and solutions

Question 1

A rigid, sealed container holds an ideal gas at an absolute pressure of 102 kPa and a temperature of \(22.0^\circ\text{C}\).

The container is heated until its temperature reaches \(82.0^\circ\text{C}\).

Calculate the final pressure of the gas.

Solution 1

The final pressure is approximately 123 kPa.

Because the container is rigid and sealed, both volume and amount of gas remain constant.

We therefore use:

\[ \frac{P_1}{T_1}=\frac{P_2}{T_2} \]

Convert the temperatures:

\[ T_1=22.0+273.15=295.15\text{ K} \]

\[ T_2=82.0+273.15=355.15\text{ K} \]

Rearrange:

\[ P_2=\frac{P_1T_2}{T_1} \]

Substitute:

\[ P_2= \frac{102\times355.15}{295.15} \]

\[ P_2=122.7\text{ kPa} \]

Therefore:

\[ \boxed{P_2=123\text{ kPa}} \]

The pressure increases because the particles move faster as the gas is heated, producing more frequent and more energetic collisions with the container walls.

Question 2

A student places a gas syringe inside a temperature-controlled water bath.

The syringe contains 285 mL of gas at \(18.0^\circ\text{C}\). The plunger moves freely, maintaining approximately constant pressure, and no gas enters or escapes.

The gas is heated to \(78.0^\circ\text{C}\).

Calculate the final gas volume.

The student instead calculates:

\[ V_2=285\times\frac{78}{18} \]

Explain why this approach is incorrect.

Solution 2

The correct final volume is approximately 344 mL, not the much larger value predicted by the student’s calculation.

Because pressure and amount remain constant, Charles’s law applies:

\[ \frac{V_1}{T_1}=\frac{V_2}{T_2} \]

First, convert both temperatures:

\[ T_1=18.0+273.15=291.15\text{ K} \]

\[ T_2=78.0+273.15=351.15\text{ K} \]

Rearrange:

\[ V_2=\frac{V_1T_2}{T_1} \]

Substitute:

\[ V_2= \frac{285\times351.15}{291.15} \]

\[ V_2=343.7\text{ mL} \]

Therefore:

\[ \boxed{V_2=344\text{ mL}} \]

The student’s mistake was using Celsius temperatures in a proportional relationship.

Their calculation assumes that volume is directly proportional to the numerical Celsius temperature.

But the correct relationship is between volume and absolute temperature.

The gas expands by approximately 20.6%, rather than increasing to more than four times its original volume.

Question 3

A chemistry research group investigates the relationship between gas pressure and temperature using a constant-volume gas thermometer.

The group repeats its measurements using two different gases.

The following are illustrative experimental observations:

• At relatively high starting gas pressures, the extrapolated pressure-temperature lines suggest zero-pressure intercepts of \(-267^\circ\text{C}\) and \(-280^\circ\text{C}\).
• When the experiments are repeated using much lower starting gas pressures, the extrapolated intercepts converge towards approximately \(-273^\circ\text{C}\).

Explain:

a) Why the experiments support the existence of a common absolute temperature scale.

b) Why the initial estimates differ between the gases.

c) Why reducing the gas pressure improves the experiment.

d) Why the experiment does not directly measure absolute zero.

Solution 3

The experiments support a common absolute temperature scale because different gases approach the same extrapolated zero-pressure temperature as their behaviour becomes more ideal.

Part a: A common absolute temperature

At constant volume and fixed amount of gas, pressure increases approximately linearly with absolute temperature.

When pressure is plotted against Celsius temperature, the relationship extrapolates towards zero pressure at approximately \(-273.15^\circ\text{C}\).

The fact that different gases approach the same intercept suggests that this temperature is not specific to the identity of the gas.

It is a fundamental temperature limit.

Part b: Why the initial estimates differ

Real gases do not behave perfectly ideally.

Their particles occupy finite volumes and experience intermolecular attractions.

Different gases have different molecular sizes and intermolecular forces, so they can deviate from ideal behaviour by different amounts.

Consequently, extrapolations made using measurements at higher gas pressures may produce different estimates.

Part c: Why lower pressure improves the experiment

At lower pressures, gas particles are further apart on average.

This reduces the influence of their intermolecular attractions and makes their individual volumes less significant compared with the total container volume.

The gases therefore behave more closely to ideal gases.

The extrapolated results become more consistent.

Part d: Why absolute zero is not measured directly

The experiment measures pressure over an accessible temperature range and extends the resulting trend mathematically.

A real gas will generally condense before reaching absolute zero.

Therefore, the experiment cannot follow a gas through the entire temperature range to directly observe zero pressure.

Instead, it uses experimental measurements and extrapolation to estimate the absolute-zero limit.

Question 4

A sealed rigid cylinder contains an ideal gas.

The following measurements are recorded:

• Volume: 2.00 L
• Temperature: \(24.0^\circ\text{C}\)
• Gauge pressure: 150 kPa
• Atmospheric pressure: 100.0 kPa

Calculate the amount of gas in the cylinder, in moles.

Explain why using 150 kPa directly in the ideal gas equation would produce an incorrect answer.

Solution 4

The cylinder contains approximately 0.202 mol of gas.

The first step is to convert gauge pressure into absolute pressure.

The ideal gas equation requires absolute pressure.

\[ P_{\text{absolute}} = P_{\text{gauge}}+P_{\text{atmospheric}} \]

\[ P_{\text{absolute}}=150+100.0 \]

\[ P_{\text{absolute}}=250.0\text{ kPa} \]

Next, convert the temperature:

\[ T=24.0+273.15 \]

\[ T=297.15\text{ K} \]

Now use:

\[ PV=nRT \]

Rearrange:

\[ n=\frac{PV}{RT} \]

Substitute the values:

\[ n= \frac{250.0\times2.00} {8.314\times297.15} \]

\[ n=0.2024\text{ mol} \]

Therefore:

\[ \boxed{n=0.202\text{ mol}} \]

Using 150 kPa directly would underestimate the amount of gas.

This is because gauge pressure measures pressure above the surrounding atmosphere, whereas the ideal gas equation requires the total pressure measured relative to a vacuum.

The temperature must also be in kelvin, independently of the pressure correction.

Question 5

A rigid, sealed vessel contains an ideal gas at \(27.0^\circ\text{C}\).

Its pressure gauge reads 200.0 kPa. Atmospheric pressure is 100.0 kPa.

The following process occurs:

1. The vessel is heated while remaining sealed until its temperature reaches \(127.0^\circ\text{C}\).
2. A valve is then opened, allowing gas to escape.
3. A heater maintains the gas at \(127.0^\circ\text{C}\) while gas escapes.
4. The valve is closed when the pressure gauge returns to its original reading of 200.0 kPa.

Assume atmospheric pressure remains constant throughout the process.

Determine the percentage of the original gas that has escaped.

A student argues that no gas has been lost overall because the final pressure reading is identical to the initial pressure reading.

Evaluate this claim.

Solution 5

25.0% of the original gas has escaped. The student’s conclusion is incorrect because identical pressure readings do not necessarily mean identical amounts of gas when the temperatures differ.

This problem combines three ideas:

• Absolute temperature.
• Absolute pressure.
• The relationship between pressure, temperature, and the amount of gas.

Step 1: Convert the initial and final gauge pressures into absolute pressures.

Initially:

\[ P_1=200.0+100.0 \]

\[ P_1=300.0\text{ kPa} \]

Finally, the gauge returns to 200.0 kPa, and atmospheric pressure is unchanged.

Therefore:

\[ P_2=300.0\text{ kPa} \]

The initial and final absolute pressures are equal.

Step 2: Convert both temperatures into kelvin.

Initial temperature:

\[ T_1=27.0+273.15 \]

\[ T_1=300.15\text{ K} \]

Final temperature:

\[ T_2=127.0+273.15 \]

\[ T_2=400.15\text{ K} \]

Step 3: Compare the initial and final gas amounts.

The ideal gas equation is:

\[ PV=nRT \]

Rearranging:

\[ n=\frac{PV}{RT} \]

Since the initial and final pressures and volumes are equal:

\[ \frac{n_2}{n_1}=\frac{T_1}{T_2} \]

Substitute:

\[ \frac{n_2}{n_1} = \frac{300.15}{400.15} \]

\[ \frac{n_2}{n_1}=0.7501 \]

Approximately 75.0% of the original gas remains.

Step 4: Calculate the percentage that escaped.

\[ \text{Fraction escaped} = 1-\frac{n_2}{n_1} \]

\[ =1-0.7501 \]

\[ =0.2499 \]

Convert to a percentage:

\[ 0.2499\times100=24.99\% \]

Therefore:

\[ \boxed{25.0\%\text{ escaped}} \]

The student’s mistake is assuming that the same pressure and volume must mean the same amount of gas.

However, the ideal gas equation contains temperature as well.

At the higher final temperature, the gas requires fewer particles to produce the same pressure inside the same container.

The final vessel has approximately the same pressure and volume as initially, but it contains 25.0% less gas.

This is why pressure alone cannot tell you whether the amount of gas has changed.

07Where to go next

When solving gas-law problems, build the habit of checking three things before choosing an equation:

1. Is temperature expressed in kelvin?
2. Is pressure absolute rather than gauge pressure?
3. Which variables remain constant, and which are changing?

Once these are clear, choosing between the ideal gas equation and the simpler gas-law relationships becomes much easier.

The next useful step is understanding how pressure, volume, and temperature appear on gas-law graphs, particularly why the temperature relationships produce straight lines through the origin when the horizontal axis is measured in kelvin.