How to Calculate Final Gas Pressure After Heating and Compression

A sealed gas sample starts at \(100.0\ \mathrm{kPa}\), occupies \(2.00\ \mathrm L\), and has a temperature of \(20.0^\circ\mathrm C\). You heat it to \(80.0^\circ\mathrm C\), while compressing it into a smaller volume of \(1.50\ \mathrm L\).

What is its final pressure?

Two things happen simultaneously. Heating the gas tends to increase its pressure, and compressing it also tends to increase its pressure. You need to account for both changes rather than applying either effect on its own.

For a fixed amount of ideal gas, use the combined gas law:

\[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \]

Rearranging to calculate final pressure:

\[ \boxed{P_2=\frac{P_1V_1T_2}{T_1V_2}} \]

The temperatures must be in kelvin, and the pressures must be absolute pressures.

Let’s work through why this equation works, how to use it correctly, and what to do when experimental measurements introduce complications.

01Why do temperature and volume both affect pressure?

Imagine gas particles moving around inside a sealed container.

They constantly collide with the container walls. These collisions produce the pressure that we measure.

Two changes can affect those collisions.

Increasing the temperature

When a gas is heated, its particles gain average kinetic energy and move faster.

They collide with the container walls more frequently and transfer more momentum during collisions. If the container’s volume remains constant, the pressure increases.

Decreasing the volume

Now imagine keeping the temperature constant but pushing a piston into the container.

The same number of particles must occupy a smaller space. They reach the walls more frequently, increasing the pressure.

Both effects can be described mathematically, but there is an important detail.

The simple relationships only work when the other relevant conditions stay constant. If temperature and volume both change, you need an equation that accounts for both.

02Deriving the combined gas law

Start with the ideal gas equation:

\[ PV=nRT \]

The symbols represent:

• \(P\): absolute pressure of the gas, in pascals (Pa) or kilopascals (kPa).
• \(V\): volume of the gas, in cubic metres (\(\mathrm{m^3}\)) or litres (L).
• \(n\): amount of gas, in moles (mol).
• \(R\): ideal gas constant.
• \(T\): absolute temperature, in kelvin (K).

Consider the same gas sample in two different states.

Initially:

\[ P_1V_1=nRT_1 \]

Finally:

\[ P_2V_2=nRT_2 \]

If the container is sealed and no gas enters or escapes, the amount of gas \(n\) stays constant.

The gas constant \(R\) also stays constant.

Rearranging both equations gives:

\[ \frac{P_1V_1}{T_1}=nR \]

\[ \frac{P_2V_2}{T_2}=nR \]

Since both expressions equal the same quantity:

\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]

This is the combined gas law.

What does the equation tell you?

The equation connects the initial and final states of the same gas sample.

Its rearranged form is particularly useful when calculating final pressure:

\[ P_2=\frac{P_1V_1T_2}{T_1V_2} \]

The subscripts tell you which measurements belong together:

A useful way to predict the direction of the pressure change is:

\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \times \frac{V_1}{V_2} \]

The temperature ratio describes the effect of heating or cooling. The volume ratio describes the effect of expansion or compression.

This gives you a quick way to check whether your calculated result makes physical sense.

For a deeper look at the individual relationships and what must remain constant, see How to Read Pressure-Volume and Temperature Gas Graphs.

03Getting the units right

Before substituting anything, check the units.

Three details cause a large proportion of mistakes in gas-pressure calculations.

Temperature must be in kelvin

The combined gas law requires absolute temperature, not degrees Celsius.

Convert using:

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

For example:

\[ 25.0^\circ\mathrm C=298.15\ \mathrm K \]

And:

\[ 100.0^\circ\mathrm C=373.15\ \mathrm K \]

Never use Celsius temperatures directly in a gas-law ratio.

Why?

The Celsius scale starts at the freezing point of water, whereas the kelvin scale starts at absolute zero. Gas-law proportionalities involving temperature depend on the absolute temperature scale.

For example, doubling the Celsius temperature from \(20^\circ\mathrm C\) to \(40^\circ\mathrm C\) does not double the absolute temperature.

The actual change is from \(293.15\ \mathrm K\) to \(313.15\ \mathrm K\).

Pressure must be absolute

Gas-law calculations use absolute pressure, measured relative to a perfect vacuum.

This is different from gauge pressure, which measures pressure relative to the surrounding atmosphere.

The relationship is:

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

For example, if a pressure gauge reads \(35\ \mathrm{kPa}\) and atmospheric pressure is \(101\ \mathrm{kPa}\):

\[ P_{\text{absolute}}=35+101 \]

\[ P_{\text{absolute}}=136\ \mathrm{kPa} \]

You must use \(136\ \mathrm{kPa}\) in the gas-law calculation, not \(35\ \mathrm{kPa}\).

This matters because gauge pressure has an arbitrary zero point: atmospheric pressure. A pressure of zero on a gauge does not mean there is no gas pressure.

Volume units must be consistent

For the combined gas law, you can use litres, millilitres, or cubic metres.

The important rule is that the initial and final volumes must use the same unit.

For example:

\[ 500\ \mathrm{mL}=0.500\ \mathrm L \]

If the initial volume is in litres and the final volume is in millilitres, convert one before substituting.

What units should you use for the gas constant?

If you use the full ideal gas equation \(PV=nRT\), your units must match the value of \(R\).

Common equivalent forms are:

These forms are equivalent because:

\[ 1\ \mathrm{kPa\,L}=1\ \mathrm J \]

The combined gas law does not require you to substitute \(R\), because it cancels when comparing two states of the same gas sample.

However, if you need to calculate the number of moles using \(PV=nRT\), choosing consistent units is essential.

05An important contrast: heating does not always increase pressure

Here is a common misconception.

A student sees that a gas has been heated and immediately predicts that its final pressure must be higher.

That prediction is only correct if the volume and amount of gas remain constant.

What happens if the gas is allowed to expand while it is heated?

The expansion reduces pressure, potentially cancelling out the effect of heating.

The combined gas law accounts for both effects.

Consider the pressure ratio:

\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \times \frac{V_1}{V_2} \]

If the temperature increases by a factor of 1.3 but the volume increases by a factor of 1.5, then:

\[ \frac{P_2}{P_1} = \frac{1.3}{1.5} \]

\[ \frac{P_2}{P_1}<1 \]

The final pressure is lower, even though the gas has been heated.

06How would you test the combined gas law experimentally?

The combined gas law is a mathematical model. An experiment can test whether the measured behaviour of a gas agrees with its predictions.

Imagine setting up a sealed gas sample inside a cylinder with a movable piston.

The apparatus contains three measuring components:

• A pressure sensor connected to the gas through a sealed side port.
• A temperature probe positioned inside the gas.
• A calibrated piston scale that measures the gas volume.

The cylinder sits inside a temperature-controlled water bath.

What should the experiment show?

First, trap a fixed amount of gas inside the cylinder.

Record its initial pressure, volume, and temperature after the gas reaches thermal equilibrium.

Next, increase the water-bath temperature and adjust the piston to produce a new volume.

Wait until the gas temperature stabilises, then record the final pressure.

The combined gas law predicts what that final pressure should be.

Consider the following illustrative measurements:

Using the combined gas law:

\[ P_2= \frac{(100.0)(1.000)(330.0)} {(300.0)(0.800)} \]

\[ P_2=137.5\ \mathrm{kPa} \]

The predicted final pressure is \(137.5\ \mathrm{kPa}\).

Suppose an illustrative experimental reading is \(137.0\pm1.0\ \mathrm{kPa}\).

The measured result agrees with the prediction within the stated measurement uncertainty.

This supports the ideal gas model under these experimental conditions. It does not prove that every real gas will behave ideally.

What could make the experiment inaccurate?

The first version of this experiment has several possible weaknesses.

1. Measuring the water-bath temperature instead of the gas temperature

The water bath may reach its target temperature before the gas does.

If you record the bath temperature too early, you may substitute an incorrect value for \(T_2\).

The improvement is to measure the gas temperature directly and wait until it stabilises.

2. Gas escaping through a connection

If the apparatus leaks, the amount of gas changes.

The combined gas law assumes a fixed amount of gas, so a leak can cause the measured pressure to disagree with the prediction.

Use airtight connections and check the apparatus before collecting measurements.

3. Piston friction

A piston that sticks may not move smoothly as the gas expands.

This can prevent the system from reaching the intended mechanical equilibrium.

A low-friction piston and repeated measurements help reduce this problem.

4. Using a gauge-pressure sensor without correcting its reading

A gauge sensor measures pressure relative to the atmosphere, not relative to a vacuum.

Convert the reading to absolute pressure before applying the gas law.

These are experimental limitations, not reasons to abandon the ideal gas equation. They show why the conditions and measuring equipment matter.

07Experimental pressure corrections and the ideal gas assumption

There is an important distinction between correcting an experimental pressure measurement and correcting the ideal gas model.

They solve different problems.

Gauge pressure versus absolute pressure

As discussed earlier, a pressure gauge measures pressure above atmospheric pressure.

The gas law requires:

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

Always make this conversion before calculating the final pressure.

Gas collected over water

In a chemistry experiment, a gas may be collected by displacing water from an inverted measuring cylinder.

The gas inside the cylinder is not necessarily pure product gas.

Some water evaporates, so the collected gas contains water vapour as well.

The total pressure is the sum of the dry gas pressure and the water vapour pressure:

\[ P_{\text{total}} = P_{\text{dry}}+P_{\ce{H2O}} \]

Rearranging:

\[ \boxed{ P_{\text{dry}} = P_{\text{total}}-P_{\ce{H2O}} } \]

This is Dalton’s law of partial pressures.

If the water levels inside and outside the measuring cylinder are equal, the total pressure of the collected gas equals atmospheric pressure.

You can then subtract the water vapour pressure to find the pressure of the dry gas.

Be careful when applying the combined gas law to gas collected over water.

If the amount of dry gas stays constant, you can apply the combined gas law to its partial pressure. However, you must account for water vapour separately if its pressure changes.

This becomes especially important when the gas temperature changes.

When does the ideal gas assumption become inaccurate?

The ideal gas equation assumes that:

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

Real gases do not satisfy these assumptions perfectly.

At relatively low pressures and high temperatures, many gases behave approximately ideally.

At high pressures and low temperatures, particle volume and intermolecular attractions become more important. The gas may also approach condensation.

For more advanced real-gas calculations, a compressibility factor \(Z\) can be introduced:

\[ PV=ZnRT \]

Here, \(Z=1\) represents ideal behaviour, while values different from 1 indicate deviations from the ideal gas model.

This is an extension beyond the routine ideal-gas calculation. Unless a question supplies a correction factor or experimental data, do not invent one.

The distinction is important:

• Adding atmospheric pressure corrects a gauge-pressure measurement.
• Subtracting water vapour pressure corrects a wet-gas measurement.
• Using a compressibility factor accounts for deviations from ideal gas behaviour.

They are three different issues.

08A reliable method for calculating final gas pressure

When you encounter a question involving simultaneous temperature and volume changes, use this sequence.

If the amount of gas changes, you cannot automatically use the ordinary combined gas law.

You may need to return to the full ideal gas equation:

\[ PV=nRT \]

For example, if a gas leaks or a chemical reaction produces additional gas, the amount of gas is no longer constant.

For a known change in the amount of gas, the more general relationship is:

\[ P_2= P_1 \left(\frac{n_2}{n_1}\right) \left(\frac{T_2}{T_1}\right) \left(\frac{V_1}{V_2}\right) \]

This is particularly useful when a question combines gas-law calculations with chemical reactions or experimental gas loss.

If you need to calculate how many moles of gas a chemical reaction produces, see Gas Volumes in Chemical Reactions for HSC Chemistry.

09Questions and solutions

Question 1

A sealed container holds an ideal gas at an absolute pressure of \(105\ \mathrm{kPa}\), a volume of \(0.750\ \mathrm L\), and a temperature of \(15.0^\circ\mathrm C\).

The gas is heated to \(45.0^\circ\mathrm C\) while its volume decreases to \(0.600\ \mathrm L\).

Calculate the final pressure.

Solution 1

The final pressure is approximately \(145\ \mathrm{kPa}\) absolute.

The amount of gas remains constant, so the combined gas law applies.

Step 1: Write down the initial and final conditions.

\[ \begin{aligned} P_1&=105\ \mathrm{kPa}\\ V_1&=0.750\ \mathrm L\\ V_2&=0.600\ \mathrm L \end{aligned} \]

Step 2: Convert temperatures to kelvin.

\[ T_1=15.0+273.15 =288.15\ \mathrm K \]

\[ T_2=45.0+273.15 =318.15\ \mathrm K \]

Step 3: Rearrange the combined gas law.

\[ P_2=\frac{P_1V_1T_2}{T_1V_2} \]

Step 4: Substitute the values.

\[ P_2= \frac{ (105)(0.750)(318.15) }{ (288.15)(0.600) } \]

\[ P_2=144.9\ \mathrm{kPa} \]

Therefore:

\[ \boxed{P_2=145\ \mathrm{kPa}} \]

The pressure increases because the gas is heated and compressed simultaneously.

Both changes increase pressure, which is consistent with the calculated result.

Question 2

A gas cylinder initially has a gauge pressure of \(42.0\ \mathrm{kPa}\). Atmospheric pressure is \(100.0\ \mathrm{kPa}\).

The gas occupies \(1.80\ \mathrm L\) at \(20.0^\circ\mathrm C\).

It is heated to \(75.0^\circ\mathrm C\) and compressed to \(1.50\ \mathrm L\).

Atmospheric pressure remains constant at \(100.0\ \mathrm{kPa}\).

Assuming ideal gas behaviour and no gas loss, calculate the final gauge pressure.

Solution 2

The final gauge pressure is approximately \(102\ \mathrm{kPa}\).

The important first step is converting the initial gauge pressure into absolute pressure.

Step 1: Convert the initial pressure.

\[ P_1=P_{\text{gauge}}+P_{\text{atm}} \]

\[ P_1=42.0+100.0 \]

\[ P_1=142.0\ \mathrm{kPa} \]

Using \(42.0\ \mathrm{kPa}\) directly in the combined gas law would be incorrect.

Step 2: Convert both temperatures.

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

\[ T_2=75.0+273.15 =348.15\ \mathrm K \]

Step 3: Apply the combined gas law.

\[ P_2=\frac{P_1V_1T_2}{T_1V_2} \]

Substitute:

\[ P_2= \frac{ (142.0)(1.80)(348.15) }{ (293.15)(1.50) } \]

\[ P_2=202.37\ \mathrm{kPa} \]

This is the final absolute pressure.

Step 4: Convert back to gauge pressure.

The question asks for the pressure displayed by the gauge, not the absolute pressure.

\[ P_{\text{gauge}}=P_{\text{absolute}}-P_{\text{atm}} \]

\[ P_{\text{gauge}}=202.37-100.0 \]

\[ \boxed{P_{\text{gauge}}=102\ \mathrm{kPa}} \]

The gas has a final absolute pressure of approximately \(202\ \mathrm{kPa}\), but the gauge displays approximately \(102\ \mathrm{kPa}\).

This distinction is essential whenever an experiment reports gauge pressure.

Question 3

A gas is collected over water at \(22.0^\circ\mathrm C\) and atmospheric pressure of \(100.8\ \mathrm{kPa}\).

The water levels inside and outside the collection cylinder are equal.

The collected gas occupies \(0.650\ \mathrm L\). The water vapour pressure at \(22.0^\circ\mathrm C\) is \(2.64\ \mathrm{kPa}\).

The water vapour is then removed without losing any dry gas. The dry gas is transferred into a sealed vessel with a volume of \(0.500\ \mathrm L\) and brought to a temperature of \(40.0^\circ\mathrm C\).

Calculate the final pressure of the dry gas, assuming ideal behaviour.

Solution 3

The final pressure of the dry gas is approximately \(135\ \mathrm{kPa}\).

The important detail is that the original gas was collected over water. Its total pressure includes the pressure contributed by water vapour.

We must correct this before using the combined gas law.

Step 1: Calculate the initial pressure of the dry gas.

Since the water levels are equal, the total pressure equals atmospheric pressure.

\[ P_{\text{total}}=100.8\ \mathrm{kPa} \]

Using Dalton’s law:

\[ P_{\text{dry}}= P_{\text{total}}-P_{\ce{H2O}} \]

\[ P_1=100.8-2.64 \]

\[ P_1=98.16\ \mathrm{kPa} \]

This is the initial pressure of the dry gas.

Step 2: Convert the temperatures to kelvin.

\[ T_1=22.0+273.15 =295.15\ \mathrm K \]

\[ T_2=40.0+273.15 =313.15\ \mathrm K \]

Step 3: Identify the initial and final volumes.

\[ V_1=0.650\ \mathrm L \]

\[ V_2=0.500\ \mathrm L \]

The dry gas is transferred without loss, so its amount remains constant.

Step 4: Apply the combined gas law.

\[ P_2=\frac{P_1V_1T_2}{T_1V_2} \]

Substitute:

\[ P_2= \frac{ (98.16)(0.650)(313.15) }{ (295.15)(0.500) } \]

\[ P_2=135.39\ \mathrm{kPa} \]

Therefore:

\[ \boxed{P_2=135\ \mathrm{kPa}} \]

The key is that the combined gas law applies to the dry gas, not to the original total pressure of the wet gas.

Using atmospheric pressure directly without subtracting the water vapour pressure would give an incorrect result.

Question 4

An ideal gas initially has an absolute pressure of \(160\ \mathrm{kPa}\), a volume of \(1.50\ \mathrm L\), and a temperature of \(27.0^\circ\mathrm C\).

The gas is heated to \(87.0^\circ\mathrm C\) and compressed to \(1.00\ \mathrm L\).

However, a leak causes 15% of the original gas to escape during the process.

Calculate the final pressure, assuming the remaining gas behaves ideally.

Solution 4

The final pressure is approximately \(245\ \mathrm{kPa}\).

This question is different from the previous ones because the amount of gas is no longer constant.

The ordinary combined gas law cannot be applied without modification.

Step 1: Determine the fraction of gas remaining.

If 15% escapes, the fraction remaining is:

\[ \frac{n_2}{n_1}=1-0.15 \]

\[ \frac{n_2}{n_1}=0.85 \]

Step 2: Convert the temperatures to kelvin.

\[ T_1=27.0+273.15 =300.15\ \mathrm K \]

\[ T_2=87.0+273.15 =360.15\ \mathrm K \]

Step 3: Use the generalised gas-law relationship.

Starting from \(PV=nRT\):

\[ P_2= P_1 \left(\frac{n_2}{n_1}\right) \left(\frac{T_2}{T_1}\right) \left(\frac{V_1}{V_2}\right) \]

Step 4: Substitute the values.

\[ P_2= 160 \times0.85 \times\frac{360.15}{300.15} \times\frac{1.50}{1.00} \]

\[ P_2=244.8\ \mathrm{kPa} \]

Therefore:

\[ \boxed{P_2=245\ \mathrm{kPa}} \]

Notice that if no gas had escaped, the final pressure would have been approximately \(288\ \mathrm{kPa}\).

The leak reduces the pressure because fewer gas particles remain in the container.

However, the pressure still increases overall because the effects of heating and compression outweigh the reduction in the amount of gas.

The tempting mistake is to assume that a sealed-sample equation still applies after a leak. The amount of gas is part of the physical situation, not a detail you can ignore.

Question 5

A sealed cylinder contains an ideal, non-reactive gas and a small quantity of liquid water.

Initially:

• The total gas pressure is \(101.3\ \mathrm{kPa}\).
• The gas volume is \(1.00\ \mathrm L\).
• The temperature is \(20.0^\circ\mathrm C\).
• The water vapour pressure is \(2.34\ \mathrm{kPa}\).

The cylinder is heated to \(50.0^\circ\mathrm C\) and compressed to a final volume of \(0.800\ \mathrm L\).

At \(50.0^\circ\mathrm C\), the saturated water vapour pressure is \(12.35\ \mathrm{kPa}\).

The following conditions apply:

• The amount of dry gas remains constant.
• Liquid water remains present throughout the process.
• The gas reaches equilibrium at both temperatures.
• The gas mixture behaves ideally.

Calculate the final total pressure inside the cylinder.

Explain why applying the ordinary combined gas law directly to the initial and final total pressures gives the wrong answer.

Solution 5

The final total pressure is approximately \(149\ \mathrm{kPa}\).

The important detail is that the amount of gas in the cylinder is not constant, even though the cylinder is sealed.

Why?

Some of the liquid water evaporates as the temperature increases, adding more water vapour to the gas mixture.

The amount of dry gas remains constant, but the total amount of gas in the gas phase changes.

This means we must calculate the dry gas pressure separately before adding the final water vapour pressure.

Step 1: Calculate the initial pressure of the dry gas.

The initial total pressure is:

\[ P_{\text{total},1}=101.3\ \mathrm{kPa} \]

The initial water vapour pressure is:

\[ P_{\ce{H2O},1}=2.34\ \mathrm{kPa} \]

Using Dalton’s law:

\[ P_{\text{dry},1} = P_{\text{total},1}-P_{\ce{H2O},1} \]

\[ P_{\text{dry},1}=101.3-2.34 \]

\[ P_{\text{dry},1}=98.96\ \mathrm{kPa} \]

Step 2: Convert the temperatures to kelvin.

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

\[ T_2=50.0+273.15 =323.15\ \mathrm K \]

Step 3: Apply the combined gas law to the dry gas only.

The amount of dry gas remains constant, so we can use:

\[ P_{\text{dry},2} = \frac{ P_{\text{dry},1}V_1T_2 }{ T_1V_2 } \]

Substitute:

\[ P_{\text{dry},2} = \frac{ (98.96)(1.00)(323.15) }{ (293.15)(0.800) } \]

\[ P_{\text{dry},2}=136.36\ \mathrm{kPa} \]

Step 4: Add the final water vapour pressure.

The question states that liquid water remains present at the final temperature.

Therefore, the gas is saturated with water vapour, and its water vapour pressure is \(12.35\ \mathrm{kPa}\).

Using Dalton’s law again:

\[ P_{\text{total},2} = P_{\text{dry},2}+P_{\ce{H2O},2} \]

\[ P_{\text{total},2} = 136.36+12.35 \]

\[ \boxed{P_{\text{total},2}=149\ \mathrm{kPa}} \]

Why does the ordinary combined gas law fail here?

A tempting approach is to apply the combined gas law directly to the initial total pressure:

\[ P_2= \frac{ (101.3)(1.00)(323.15) }{ (293.15)(0.800) } \]

This gives:

\[ P_2=139.6\ \mathrm{kPa} \]

That answer is incorrect.

The combined gas law assumes that the amount of gas stays constant. In this question, some liquid water evaporates as the temperature increases.

The number of water vapour particles in the gas phase therefore increases.

The dry gas follows the combined gas law because its amount remains constant. The water vapour does not follow that same calculation because its amount changes.

The final pressure must account for both components.

This is the conceptual distinction that makes the question difficult: a sealed container does not necessarily contain a constant amount of gas in its gas phase when a liquid can evaporate inside it.

The most important decision in any final-pressure calculation is identifying which gas particles remain in the sample and which can enter or leave the gas phase. Once you understand that, you can choose between the combined gas law, the full ideal gas equation, and partial-pressure corrections without confusing their assumptions.