Why Gas Volume Increases When Temperature Rises at Constant Pressure

Put a gas syringe in warm water and watch its plunger move out. The gas occupies more space, even though the pressure pushing against the plunger stays constant.

Why doesn’t the pressure increase when the gas gets hotter? After all, its particles are moving faster and colliding with the walls more energetically.

The answer is that the gas is allowed to expand. As its temperature increases, the same number of particles spreads through a larger volume, keeping the pressure constant. For a fixed amount of ideal gas at constant pressure, volume is directly proportional to absolute temperature.

The particles themselves do not get bigger. The space between them increases.

01Why does heating make a gas expand?

Start with the particles.

A gas contains particles moving randomly in every direction. They collide with the walls of their container, transferring momentum. These collisions produce the pressure we measure.

When you heat a gas, you increase the average kinetic energy of its particles. They move faster, and their collisions with the container walls become more energetic.

What happens next depends on whether the container can expand.

What happens when the container can move?

Imagine gas trapped inside a cylinder with a movable piston.

The piston can slide upwards, but a fixed weight pushes down on it. Atmospheric pressure also acts on the piston.

Initially, the gas pressure balances the external pressure.

Now heat the gas.

1. The particles gain kinetic energy and move faster.
2. Their more energetic collisions initially tend to increase the pressure.
3. The gas pushes the piston upwards.
4. The volume increases, spreading the same number of particles through more space.
5. The reduced particle concentration counteracts the increased collision energy.

The piston eventually reaches a new equilibrium where the gas pressure again matches the external pressure.

The gas is now hotter and occupies a larger volume, but its pressure has not changed.

This is the key idea: at constant pressure, heating allows a gas to expand rather than maintain its original volume and develop a higher pressure.

02How do we test this experimentally?

To investigate the relationship, we need an apparatus that allows gas volume to change while keeping pressure approximately constant.

A useful arrangement consists of:

• A graduated cylinder containing a fixed amount of gas.
• A freely moving, low-friction piston.
• A constant mass placed on the piston.
• A water bath to heat the gas gradually.
• A thermometer to measure the temperature.

Why does the apparatus need a movable piston?

Consider a different experimental design.

Suppose we trap gas inside a sealed, rigid flask and heat it.

The volume cannot increase because the flask cannot expand. Instead, the pressure rises.

This is useful for investigating how pressure changes with temperature, but it cannot directly demonstrate how gas volume changes at constant pressure.

We therefore replace the rigid container with a movable piston.

The piston allows the volume to change while the external pressure remains approximately constant.

How do we keep the pressure constant?

The pressure acting on the gas is determined by atmospheric pressure and the load on the piston.

For a frictionless piston moving slowly enough that its acceleration is negligible:

\[ P_{\mathrm{gas}}=P_{\mathrm{atm}}+\frac{mg}{A} \]

where:

• \(P_{\mathrm{gas}}\) is the absolute pressure of the gas.
• \(P_{\mathrm{atm}}\) is atmospheric pressure.
• \(m\) is the total moving mass supported by the piston.
• \(g\) is gravitational acceleration, approximately \(9.81\ \mathrm{m\,s^{-2}}\).
• \(A\) is the cross-sectional area of the piston.

If atmospheric pressure, the mass, and the piston area remain unchanged, the external pressure remains constant.

As the gas heats up, the piston moves upwards. We measure the new volume using the graduated cylinder.

The experiment should show that increasing the absolute temperature produces a proportional increase in volume.

What should the measurements look like?

Consider the following illustrative experimental data for a fixed amount of gas under constant pressure.

Illustrative measurements, not results from an actual experiment.

Notice that the ratio of volume to absolute temperature remains approximately constant.

For example, using the first measurement:

\[ \frac{V}{T}=\frac{100.0}{293.15} \]

\[ \frac{V}{T}=0.3411\ \mathrm{mL\,K^{-1}} \]

Using the final measurement:

\[ \frac{V}{T}=\frac{120.4}{353.15} \]

\[ \frac{V}{T}=0.3409\ \mathrm{mL\,K^{-1}} \]

These values are very close.

Small differences are expected because experimental measurements have limited precision. The results support the predicted proportional relationship.

03From the particle model to Charles’s law

We can now express the relationship mathematically.

Start with the ideal gas equation:

\[ PV=nRT \]

where:

• \(P\) is the absolute pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the amount of gas in moles.
• \(R\) is the ideal gas constant.
• \(T\) is the absolute temperature in kelvin.

For a fixed amount of gas at constant pressure, \(P\) and \(n\) do not change.

Rearranging:

\[ \frac{V}{T}=\frac{nR}{P} \]

Since \(n\), \(R\), and \(P\) are constant, the right-hand side remains constant.

Therefore:

\[ \boxed{\frac{V}{T}=\text{constant}} \]

This is Charles’s law.

It can also be written as:

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

where:

• \(V_1\) is the initial volume.
• \(T_1\) is the initial absolute temperature.
• \(V_2\) is the final volume.
• \(T_2\) is the final absolute temperature.

The subscripts 1 and 2 simply identify the initial and final states.

Why must temperature be in kelvin?

This is one of the most important details in gas-law calculations.

Charles’s law describes a directly proportional relationship between volume and absolute temperature.

Celsius does not measure temperature from absolute zero. Its zero point is based on the freezing point of water.

Kelvin starts at absolute zero, the lowest possible thermodynamic temperature.

To convert Celsius to kelvin:

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

Here, \(t\) is the temperature in degrees Celsius and \(T\) is the temperature in kelvin.

For example:

A temperature increase of \(20^\circ\mathrm C\) is also an increase of 20 K. However, temperature ratios must use kelvin.

Imagine heating a gas from 20°C to 40°C.

A tempting prediction is that doubling the Celsius temperature doubles the volume.

That is incorrect.

The actual temperatures are 293.15 K and 313.15 K. The volume increases by only about 6.8%, not 100%.

The reason is that Celsius zero is an arbitrary reference point for the gas-law calculation.

04What units should you use for pressure, temperature, and R?

You will often need the ideal gas equation rather than just Charles’s law.

The units must be consistent.

The ideal gas constant is:

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

Equivalent forms are:

\[ R=8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}} \]

or:

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

The last form is particularly convenient for HSC Chemistry calculations involving pressure in kPa and volume in litres.

Remember that \(1\ \mathrm{kPa\,L}=1\ \mathrm J\).

For a more detailed explanation of choosing the correct gas constant, see Choosing R When Pressure Is in kPa and Volume Is in L.

One more important detail: the pressure in \(PV=nRT\) is absolute pressure, not gauge pressure.

Gauge pressure measures pressure relative to atmospheric pressure. A gauge reading of zero does not mean there is no pressure. It means the gas pressure equals atmospheric pressure.

07What if the pressure is not constant?

Here is the important contrasting situation.

Suppose the same gas is trapped inside a sealed, rigid steel container.

You heat it, but the container cannot expand.

The gas particles still gain kinetic energy, but the volume cannot increase. Instead, the pressure rises.

This is a different gas-law relationship.

At constant volume and constant amount of gas:

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

Compare the two situations:

The important lesson is that heating does not automatically increase gas volume.

The result depends on which physical conditions are allowed to change.

08Ideal gas assumptions versus experimental corrections

Charles’s law is derived using the ideal gas model. However, real gases and real laboratory equipment are not perfectly ideal.

It is important to distinguish two different sources of disagreement between a prediction and an experimental result.

1. Limitations of the ideal gas model

An ideal gas assumes:

• Gas particles occupy negligible volume compared with the container.
• There are no significant intermolecular attractions or repulsions.
• Collisions between particles and with the container walls are perfectly elastic.
• The particles move randomly.

Real gas particles have finite sizes and experience intermolecular forces.

At high pressures, particles are closer together, so their own volume becomes more significant. At low temperatures, intermolecular attractions become more important.

Consequently, a real gas may not follow the ideal gas equation perfectly.

At typical conditions where gases behave approximately ideally, Charles’s law provides a useful prediction.

At more extreme conditions, deviations from ideal behaviour may become significant. More advanced real-gas models introduce corrections for molecular volume and intermolecular attractions.

These are corrections to the theoretical model, not corrections to a faulty measurement.

2. Experimental errors and corrections

Even if the ideal gas model is appropriate, the experimental apparatus can affect the measurements.

One particularly important experimental issue is piston friction.

If the piston sticks while the gas is heated, the gas pressure can increase temporarily. When the pressure becomes large enough to overcome friction, the piston suddenly moves.

The experiment may then appear to show that pressure increases during heating, even though the intended investigation is at constant pressure.

This is why a low-friction piston and slow heating are important.

Also remember that a real-gas deviation is not automatically evidence of experimental error. The model and the measurement method must be evaluated separately.

09Questions and solutions

Question 1

A fixed amount of ideal gas occupies 180 mL at 17.0°C.

The gas is heated to 67.0°C while the pressure remains constant.

Calculate its final volume.

Solution 1

The final volume is approximately 211 mL.

Since the amount of gas and pressure remain constant, Charles’s law applies.

Step 1: Convert both temperatures into kelvin.

\[ T_1=17.0+273.15=290.15\ \mathrm K \]

\[ T_2=67.0+273.15=340.15\ \mathrm K \]

Step 2: Apply Charles’s law.

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

Rearrange:

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

Substitute:

\[ \begin{aligned} V_2&=\frac{180\times340.15}{290.15}\\ &=211.0\ \mathrm{mL} \end{aligned} \]

Therefore:

\[ \boxed{V_2=211\ \mathrm{mL}} \]

The gas expands because its absolute temperature increases while its pressure and amount remain constant.

Question 2

An ideal gas is contained in a vertical cylinder with a freely moving piston. The piston is exposed to atmospheric pressure, and there is no additional load. Assume the piston has negligible mass and friction.

A pressure gauge reads 0.0 kPa gauge.

Initially, the gas occupies 0.750 L at 20.0°C. It is heated to 120.0°C.

Assume atmospheric pressure remains constant at 101.3 kPa and no gas escapes.

Calculate the final volume. Explain why the gauge reading does not mean the gas pressure is zero.

Solution 2

The final volume is approximately 1.01 L. The gas pressure is not zero; it is equal to atmospheric pressure.

Step 1: Identify the pressure condition.

A gauge measures pressure relative to atmospheric pressure.

A gauge reading of zero therefore means:

\[ P_{\mathrm{gas}}=P_{\mathrm{atm}} \]

The absolute gas pressure is:

\[ P_{\mathrm{gas}}=101.3\ \mathrm{kPa} \]

The pressure remains constant because the piston moves freely against the same atmospheric pressure.

Step 2: Convert the temperatures.

\[ T_1=20.0+273.15=293.15\ \mathrm K \]

\[ T_2=120.0+273.15=393.15\ \mathrm K \]

Step 3: Apply Charles’s law.

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

\[ \begin{aligned} V_2&=\frac{(0.750)(393.15)}{293.15}\\ &=1.006\ \mathrm L \end{aligned} \]

Therefore:

\[ \boxed{V_2=1.01\ \mathrm L} \]

The gas expands because the absolute temperature increases.

The trap is confusing gauge pressure with absolute pressure. A zero gauge reading does not mean a vacuum exists.

Question 3

A student investigates the relationship between gas volume and temperature using a sealed sample of gas beneath a freely moving piston.

The external pressure is kept constant.

The following illustrative measurements are obtained:

The volume uncertainty is ±0.5 mL, and temperature uncertainty is negligible.

(a) Use the first measurement and Charles’s law to predict the volume at 65.0°C.

(b) Calculate \(V/T\) for the first and last measurements. Explain whether the results support Charles’s law.

(c) Suggest one experimental reason why the ratios are not perfectly identical.

Solution 3

(a) The predicted volume at 65.0°C is approximately 115.4 mL.

The gas amount and pressure are constant, so:

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

Convert the temperatures:

\[ T_1=20.0+273.15=293.15\ \mathrm K \]

\[ T_2=65.0+273.15=338.15\ \mathrm K \]

Substitute:

\[ \begin{aligned} V_2&=\frac{100.0\times338.15}{293.15}\\ &=115.35\ \mathrm{mL} \end{aligned} \]

Therefore:

\[ \boxed{V_2=115.4\ \mathrm{mL}} \]

(b) The results support Charles’s law.

First, calculate the ratio for the initial measurement:

\[ \frac{V_1}{T_1}=\frac{100.0}{293.15} \]

\[ \frac{V_1}{T_1}=0.3411\ \mathrm{mL\,K^{-1}} \]

Now calculate the ratio for the final measurement:

\[ \frac{V_2}{T_2}=\frac{120.4}{353.15} \]

\[ \frac{V_2}{T_2}=0.3409\ \mathrm{mL\,K^{-1}} \]

The ratios are very close.

Their small difference is consistent with the stated measurement uncertainty, so the results support the predicted direct proportionality.

(c) One possible reason is piston friction.

If the piston experiences friction, it may not move smoothly as the gas expands. This can cause small differences between the actual gas pressure and the assumed constant external pressure.

Other possibilities include reading uncertainty or the gas not reaching thermal equilibrium with the water bath.

Question 4

Two identical samples of an ideal gas initially occupy 1.50 L at 25.0°C and an absolute pressure of 100.0 kPa.

Both samples are heated to 125.0°C.

However, they are placed in different containers:

• Container A is a sealed, rigid steel vessel.
• Container B is a sealed cylinder with a freely moving piston under a constant external load.

Assume both samples behave ideally and have no gas leakage.

(a) Calculate the final pressure in container A.

(b) Calculate the final volume in container B.

(c) Explain why the two samples respond differently despite undergoing the same temperature change.

Solution 4

(a) The final pressure in container A is approximately 133.5 kPa.

(b) The final volume in container B is approximately 2.00 L.

(c) The difference occurs because the containers impose different physical constraints.

First, convert the temperatures:

\[ T_1=25.0+273.15=298.15\ \mathrm K \]

\[ T_2=125.0+273.15=398.15\ \mathrm K \]

Part (a): Rigid container

The volume cannot change, and the amount of gas remains constant.

Therefore, use the pressure-temperature relationship:

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

Rearrange:

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

Substitute:

\[ \begin{aligned} P_2&=\frac{100.0\times398.15}{298.15}\\ &=133.5\ \mathrm{kPa} \end{aligned} \]

The final pressure is:

\[ \boxed{P_2=133.5\ \mathrm{kPa}} \]

The pressure increases because the gas cannot expand.

Part (b): Movable piston

The pressure remains constant because the piston can move against a fixed external load.

Therefore, use Charles’s law:

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

Rearrange:

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

Substitute:

\[ \begin{aligned} V_2&=\frac{1.50\times398.15}{298.15}\\ &=2.003\ \mathrm L \end{aligned} \]

Therefore:

\[ \boxed{V_2=2.00\ \mathrm L} \]

Part (c): Why are the results different?

Both samples experience the same temperature increase, but their containers impose different constraints.

In container A, the rigid walls prevent expansion, so pressure increases.

In container B, the piston moves outwards, allowing volume to increase while pressure remains constant.

The temperature change alone does not determine whether pressure or volume increases. The experimental conditions determine which variable can change.

Question 5

A sealed cylinder contains an ideal gas beneath a frictionless, movable piston.

Initially, the gas has:

\[ P_0=100.0\ \mathrm{kPa} \]

\[ V_0=0.600\ \mathrm L \]

\[ T_0=300\ \mathrm K \]

The piston moves freely against a constant external pressure of 100.0 kPa. However, a rigid upper stop prevents the piston from expanding beyond a volume of 0.900 L.

The following procedure is carried out:

1. The gas is heated slowly until the piston reaches the upper stop.
2. The piston is latched in place, and the gas is heated further to 600 K.
3. While the piston remains locked at 600 K, a valve is opened until exactly 20.0% of the original gas particles have escaped. The valve is then closed.
4. The gas is cooled to 300 K while the piston remains locked.
5. The latch is released. The external pressure remains at 100.0 kPa, and no more gas escapes.

Determine:

(a) The temperature at which the piston first reaches the upper stop.

(b) The gas pressure after the gas has been heated to 600 K while the piston is locked.

(c) The gas pressure after 20.0% of the particles have escaped.

(d) The final volume after the gas has cooled to 300 K and the piston has been released.

Assume the gas behaves ideally throughout, the piston is frictionless, and all temperatures refer to the gas itself.

Solution 5

The final volume is 0.480 L.

The crucial detail is that three different conditions occur during the procedure:

• The gas initially expands at constant pressure.
• The gas is then heated at constant volume.
• The amount of gas decreases before the piston is released.

Applying Charles’s law throughout the entire process would give the wrong answer because the pressure and amount of gas do not remain constant throughout.

Step 1: Calculate the temperature when the piston reaches the stop.

Initially, the piston moves freely, so pressure and amount of gas remain constant.

Charles’s law applies:

\[ \frac{V_0}{T_0}=\frac{V_{\mathrm{stop}}}{T_{\mathrm{stop}}} \]

Rearrange:

\[ T_{\mathrm{stop}}= \frac{T_0V_{\mathrm{stop}}}{V_0} \]

Substitute:

\[ \begin{aligned} T_{\mathrm{stop}} &=\frac{300\times0.900}{0.600}\\ &=450\ \mathrm K \end{aligned} \]

Therefore:

\[ \boxed{T_{\mathrm{stop}}=450\ \mathrm K} \]

The piston reaches its mechanical limit at 450 K.

Step 2: Calculate the pressure after heating to 600 K.

The piston is now locked.

The volume and amount of gas remain constant during this stage.

We therefore use:

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

At the start of this stage:

\[ P_1=100.0\ \mathrm{kPa} \]

\[ T_1=450\ \mathrm K \]

At the end:

\[ T_2=600\ \mathrm K \]

Substitute:

\[ \begin{aligned} P_2&=\frac{100.0\times600}{450}\\ &=133.3\ \mathrm{kPa} \end{aligned} \]

Therefore:

\[ \boxed{P_2=133.3\ \mathrm{kPa}} \]

The pressure rises because the piston cannot move any further.

Step 3: Account for the gas escaping at 600 K.

The valve releases exactly 20.0% of the original gas particles.

Therefore, 80.0% remain:

\[ n_3=0.800n_0 \]

The temperature and volume are unchanged while the gas escapes.

From the ideal gas equation:

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

At constant temperature and volume, pressure is directly proportional to the amount of gas.

Therefore:

\[ P_3=0.800P_2 \]

Substitute:

\[ \begin{aligned} P_3&=0.800\times133.3\\ &=106.7\ \mathrm{kPa} \end{aligned} \]

Thus:

\[ \boxed{P_3=106.7\ \mathrm{kPa}} \]

The pressure decreases because some gas particles have escaped.

Step 4: Calculate the pressure after cooling to 300 K.

The piston remains locked, so volume stays constant.

No further gas escapes, so the amount of gas also remains constant.

We can use the pressure-temperature relationship:

\[ \frac{P_3}{T_3}=\frac{P_4}{T_4} \]

The gas cools from 600 K to 300 K.

Therefore:

\[ \begin{aligned} P_4&=\frac{P_3T_4}{T_3}\\ &=\frac{106.7\times300}{600}\\ &=53.3\ \mathrm{kPa} \end{aligned} \]

The pressure is now:

\[ \boxed{P_4=53.3\ \mathrm{kPa}} \]

This is below the external pressure of 100.0 kPa.

Consequently, when the latch is released, the external pressure pushes the piston downwards, compressing the remaining gas.

Step 5: Calculate the final volume.

After the piston is released, the gas returns to mechanical equilibrium with the external pressure.

The final conditions are:

\[ P_f=100.0\ \mathrm{kPa} \]

\[ T_f=300\ \mathrm K \]

However, only 80.0% of the original gas remains.

Apply the ideal gas equation to compare the initial and final states:

\[ \frac{P_0V_0}{n_0T_0} = \frac{P_fV_f}{n_fT_f} \]

Rearranging:

\[ V_f=V_0 \left(\frac{n_f}{n_0}\right) \left(\frac{T_f}{T_0}\right) \left(\frac{P_0}{P_f}\right) \]

Substitute:

\[ \begin{aligned} V_f&=0.600 \times0.800 \times\frac{300}{300} \times\frac{100.0}{100.0}\\ &=0.480\ \mathrm L \end{aligned} \]

Therefore:

\[ \boxed{V_f=0.480\ \mathrm L} \]

Why is this result surprising?

There are two tempting mistakes.

The first is to apply Charles’s law throughout the entire heating process:

\[ V_2=0.600\times\frac{600}{300} =1.20\ \mathrm L \]

But this ignores the mechanical stop at 0.900 L. Once the piston reaches that limit, the volume cannot increase, so pressure rises instead.

The second mistake is to assume that returning to the original temperature and pressure must restore the original volume.

That would only be true if the amount of gas also remained unchanged.

Here, 20.0% of the original particles escaped. At the original temperature and pressure, the remaining gas must occupy 80.0% of the original volume.

That is why the final volume is 0.480 L rather than 0.600 L.

The next useful step is to connect these physical conditions to gas-law graphs. Before deciding whether a graph should show increasing pressure or increasing volume, identify what the experiment actually holds constant. For the mathematical graph relationships, see How to Read Pressure-Volume and Temperature Gas Graphs.