How to Interpret a Pressure vs Inverse-Volume Graph

A sealed syringe contains a fixed amount of gas. You slowly compress it until its volume halves, keeping its temperature constant. What happens to the pressure?

It doubles.

Now suppose you plot pressure against volume. You get a decreasing curve. But if you plot pressure against the inverse of volume, something much more useful happens: the graph becomes a straight line.

This is the key to interpreting a pressure vs inverse-volume graph. Its gradient tells you something about the amount of gas and its temperature, while its intercept can reveal whether the experiment has been set up or measured correctly.

01Why does plotting inverse volume produce a straight line?

Start with what happens when you compress a gas.

The same gas particles are confined to a smaller space. They collide with the container walls more frequently, increasing the pressure.

If the temperature and amount of gas remain constant, Boyle’s law describes this relationship:

\[ PV=k \]

Here:

• \(P\) is the absolute pressure of the gas.
• \(V\) is its volume.
• \(k\) is a constant for that particular sample at that temperature.

Rearranging:

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

This tells us that pressure is inversely proportional to volume.

For example:

• Double the volume and the pressure halves.
• Halve the volume and the pressure doubles.
• Triple the volume and the pressure becomes one-third of its original value.

The graph of pressure against volume is therefore a decreasing curve.

What changes when we use inverse volume?

Instead of plotting \(P\) against \(V\), we plot \(P\) against \(1/V\).

The relationship becomes:

\[ P=k\left(\frac{1}{V}\right) \]

This has the form of a straight-line equation:

\[ y=mx \]

where:

• \(y=P\), the pressure.
• \(x=1/V\), the inverse volume.
• \(m=k\), the gradient.

Therefore, a graph of pressure against inverse volume should be a straight line through the origin, provided the amount of gas and temperature remain constant and the gas behaves ideally.

The important distinction: pressure is inversely proportional to volume, but directly proportional to inverse volume.

These are two ways of describing the same relationship.

02What does the gradient actually represent?

The ideal gas equation gives us a more useful interpretation of the gradient.

\[ PV=nRT \]

Where:

Rearrange the equation to make pressure the subject:

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

Or:

\[ P=nRT\left(\frac{1}{V}\right) \]

Compare this with the straight-line equation:

\[ y=mx \]

We can identify:

\[ \boxed{\text{Gradient}=nRT} \]

This is the most important result when interpreting a pressure vs inverse-volume graph.

The gradient is not simply \(R\). It is the product of the amount of gas, the gas constant, and the absolute temperature.

What units should you use?

For pressure in kilopascals and volume in litres, use:

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

The corresponding graph units are:

• Vertical axis: pressure in kPa.
• Horizontal axis: inverse volume in L⁻¹.
• Gradient: kPa L.

The units are consistent because:

\[ \frac{\mathrm{kPa}}{\mathrm{L^{-1}}} =\mathrm{kPa\,L} \]

You can also use pressure in pascals and volume in cubic metres, with:

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

These are equivalent combinations because \(1\ \mathrm{kPa\,L}=1\ \mathrm{J}\).

Temperature must always be expressed in kelvin, not degrees Celsius.

For example:

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

If your graph uses inverse volume in \(\mathrm{cm^{-3}}\), its gradient will be in \(\mathrm{kPa\,cm^3}\). Convert the units or use a matching value of \(R\) before calculating the amount of gas.

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

03How is this relationship tested experimentally?

Imagine you want to investigate whether pressure is directly proportional to inverse volume.

You need to change the gas volume while keeping its temperature and amount constant.

A suitable experimental arrangement uses a gas syringe connected to a pressure sensor.

The experiment works as follows:

1. Introduce a known amount of dry gas into the syringe and seal the apparatus.
2. Connect the syringe to the pressure sensor.
3. Place the syringe in a temperature-controlled water bath.
4. Allow the gas to reach the bath temperature.
5. Slowly move the plunger to change the volume.
6. Record the pressure and corresponding gas volume at several positions.
7. Calculate the reciprocal of each volume and plot pressure against inverse volume.

The volume must include all the space occupied by the gas, including any connecting tubes and the pressure sensor cavity if their volumes are significant.

What should the experiment show?

Suppose the sample contains \(0.0500\ \mathrm{mol}\) of gas at \(25.0^\circ\mathrm C\).

The temperature is:

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

The ideal gas equation predicts a gradient of:

\[ \begin{aligned} m&=nRT\\ &=0.0500\times8.314\times298.15\\ &\approx123.9\ \mathrm{kPa\,L} \end{aligned} \]

The following readings are illustrative data, not measurements from a real experiment.

Notice what happens as the volume decreases.

The inverse volume increases, and the pressure increases in the same proportion.

The plotted points should lie approximately on a straight line through the origin.

Why might the first experimental attempt fail?

Imagine compressing the syringe quickly and recording the pressure immediately.

You might find that the measured pressure is higher than expected.

Does this mean Boyle’s law is wrong?

Not necessarily.

Rapid compression heats the gas. Its temperature is no longer constant, so the experiment has changed two variables instead of just one.

The solution is to compress the gas slowly and allow it to return to the bath temperature before recording each measurement.

This is an important experimental distinction: a straight-line relationship predicted by a model only applies when the required conditions are maintained.

06What does the y-intercept tell you?

For an ideal gas, the equation is:

\[ P=nRT\left(\frac{1}{V}\right) \]

There is no additional constant term.

Therefore, the graph should have a y-intercept of zero.

However, experimental graphs do not always pass through the origin.

This does not automatically mean that the gas is behaving non-ideally or that Boyle’s law has failed.

The pressure measurement itself may need correcting.

Case 1: The sensor measures gauge pressure

There are two ways to measure pressure.

• Absolute pressure is measured relative to a perfect vacuum.
• Gauge pressure is measured relative to atmospheric pressure.

The ideal gas equation requires absolute pressure.

The relationship is:

\[ P_{\mathrm{abs}}=P_{\mathrm{gauge}}+P_{\mathrm{atm}} \]

Rearranging the ideal gas equation in terms of gauge pressure:

\[ P_{\mathrm{gauge}} =nRT\left(\frac{1}{V}\right)-P_{\mathrm{atm}} \]

This equation still describes a straight line, but its y-intercept is negative.

For example, if atmospheric pressure is \(101.3\ \mathrm{kPa}\), the gauge-pressure graph has a y-intercept of \(-101.3\ \mathrm{kPa}\).

The gradient remains \(nRT\).

Case 2: The gas contains water vapour

Suppose a gas is collected over water.

The pressure sensor measures the combined pressure exerted by the dry gas and the water vapour.

According to Dalton’s law of partial pressures:

\[ P_{\mathrm{total}}=P_{\mathrm{dry}}+P_{\mathrm{H_2O}} \]

Therefore:

\[ P_{\mathrm{total}} =nRT\left(\frac{1}{V}\right)+P_{\mathrm{H_2O}} \]

If the gas is saturated with water vapour and the temperature remains constant, the water vapour pressure remains constant.

This produces a positive y-intercept equal to the water vapour pressure.

To calculate the pressure of the dry gas, subtract the water vapour pressure:

\[ P_{\mathrm{dry}} =P_{\mathrm{total}}-P_{\mathrm{H_2O}} \]

Compare the three graphs

A pressure sensor can also have a zero offset, producing another non-zero intercept.

The key is to investigate the pressure reference and measurement conditions before deciding that the ideal gas relationship has failed.

07Ideal gas assumptions vs experimental corrections

It is important to distinguish between an imperfect experiment and an imperfect model.

The ideal gas equation assumes that gas particles have negligible individual volume and that intermolecular attractions and repulsions are insignificant.

Real gases do not satisfy these assumptions perfectly.

Deviations become more significant at high pressures and low temperatures.

However, several apparent deviations are actually caused by experimental problems.

One particularly important correction is the total gas volume.

If a syringe is connected to a pressure sensor, some gas occupies the connecting tube and sensor cavity.

Therefore:

\[ V_{\mathrm{total}} =V_{\mathrm{syringe}}+V_{\mathrm{dead}} \]

Using only the syringe reading gives an incorrect value for \(1/V\).

This can distort the graph, particularly when the syringe volume becomes small.

Remember that experimental corrections do not change the universal gas constant. They correct the quantities being measured.

08A reliable method for interpreting an exam graph

When you are given a pressure vs inverse-volume graph, use this sequence.

Step 1: Check the axes.

Identify pressure on the vertical axis and inverse volume on the horizontal axis.

Pay attention to the reciprocal-volume units.

Step 2: Check the experimental conditions.

Confirm that the amount of gas and temperature are constant.

If either changes, the gradient may change.

Step 3: Identify the expected relationship.

For an ideal gas:

\[ P=nRT\left(\frac{1}{V}\right) \]

The expected graph is a straight line with a positive gradient.

Step 4: Interpret the gradient.

Remember:

\[ \boxed{m=nRT} \]

You can use this to calculate the amount of gas if the temperature is known.

Step 5: Check the intercept.

For absolute pressure of a dry ideal gas, the expected intercept is zero.

If it is not zero, investigate gauge pressure, water vapour, sensor calibration, and volume corrections.

Step 6: Check the units before calculating.

Use kelvin for temperature and compatible pressure, volume, and gas constant units.

For a broader comparison of the different gas-law graphs and the conditions that produce them, see How to Read Pressure-Volume and Temperature Gas Graphs.

09Questions and solutions

Question 1

A fixed amount of ideal gas is maintained at constant temperature.

Its initial absolute pressure is \(96.0\ \mathrm{kPa}\), and its volume is \(0.750\ \mathrm L\).

The gas is compressed to \(0.500\ \mathrm L\).

Calculate:

a) The final pressure.

b) The initial and final inverse-volume coordinates.

Solution 1

The final pressure is \(144\ \mathrm{kPa}\), and the inverse-volume coordinate increases from \(1.33\ \mathrm{L^{-1}}\) to \(2.00\ \mathrm{L^{-1}}\).

Since the amount and temperature remain constant, Boyle’s law applies:

\[ P_1V_1=P_2V_2 \]

Rearrange:

\[ P_2=\frac{P_1V_1}{V_2} \]

Substitute:

\[ \begin{aligned} P_2&=\frac{96.0\times0.750}{0.500}\\ &=144\ \mathrm{kPa} \end{aligned} \]

Now calculate the inverse volumes.

Initially:

\[ \frac{1}{V_1}=\frac{1}{0.750} =1.33\ \mathrm{L^{-1}} \]

Finally:

\[ \frac{1}{V_2}=\frac{1}{0.500} =2.00\ \mathrm{L^{-1}} \]

The point moves upwards and to the right on the pressure vs inverse-volume graph.

This is consistent with a positive-gradient straight line.

Question 2

A sealed sample of ideal gas is investigated at two different temperatures.

The following gradients are obtained from pressure vs inverse-volume graphs:

Determine the amount of gas calculated from each experiment.

Are the results consistent with the same sample being used? Explain why the gradients differ.

Solution 2

Both experiments give approximately \(0.0484\ \mathrm{mol}\), so the results are consistent with the same amount of gas.

The gradient relationship is:

\[ m=nRT \]

Rearrange:

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

For experiment A:

\[ \begin{aligned} n_A&=\frac{120.0}{8.314\times298.0}\\ &=\frac{120.0}{2477.572}\\ &=0.0484\ \mathrm{mol} \end{aligned} \]

For experiment B:

\[ \begin{aligned} n_B&=\frac{141.0}{8.314\times350.0}\\ &=\frac{141.0}{2909.9}\\ &=0.0485\ \mathrm{mol} \end{aligned} \]

The small difference is due to rounding.

Both results are approximately \(0.0484\ \mathrm{mol}\).

Why is the second gradient steeper?

Because the temperature is higher.

For a fixed amount of gas:

\[ m\propto T \]

The gradient increases in proportion to absolute temperature.

The tempting mistake is to assume that a steeper gradient must indicate more gas. However, the gradient depends on both \(n\) and \(T\).

Question 3

A student collects a dry gas over water at \(25.0^\circ\mathrm C\).

The gas is saturated with water vapour, which has a pressure of \(3.17\ \mathrm{kPa}\) at this temperature.

An absolute-pressure sensor measures the total pressure.

The graph of total pressure against inverse volume is described by:

\[ P_{\mathrm{total}} =82.0\left(\frac{1}{V}\right)+3.17 \]

Here, pressure is in kPa and volume is in L.

Calculate:

a) The amount of dry gas present.

b) The total pressure when the volume is \(0.500\ \mathrm L\).

c) The apparent amount of gas if the student incorrectly uses the total pressure directly in the ideal gas equation.

Solution 3

The dry gas amount is approximately \(0.0331\ \mathrm{mol}\), and the total pressure at \(0.500\ \mathrm L\) is \(167.17\ \mathrm{kPa}\). Using total pressure without correction overestimates the amount of dry gas.

The equation is:

\[ P_{\mathrm{total}} =nRT\left(\frac{1}{V}\right)+P_{\mathrm{H_2O}} \]

The gradient is \(82.0\ \mathrm{kPa\,L}\).

Therefore:

\[ m=nRT \]

Part a: Calculate the amount of dry gas.

First, convert the temperature:

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

Rearrange:

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

Substitute:

\[ \begin{aligned} n&=\frac{82.0}{8.314\times298.15}\\ &=\frac{82.0}{2478.82}\\ &=0.0331\ \mathrm{mol} \end{aligned} \]

Part b: Calculate the total pressure.

At \(V=0.500\ \mathrm L\):

\[ \frac{1}{V}=2.00\ \mathrm{L^{-1}} \]

Substitute into the graph equation:

\[ \begin{aligned} P_{\mathrm{total}} &=82.0\times2.00+3.17\\ &=167.17\ \mathrm{kPa} \end{aligned} \]

Part c: Calculate the incorrect apparent amount.

If the student uses the total pressure directly:

\[ n_{\mathrm{apparent}}=\frac{PV}{RT} \]

Substitute:

\[ \begin{aligned} n_{\mathrm{apparent}} &=\frac{167.17\times0.500} {8.314\times298.15}\\ &=\frac{83.585}{2478.82}\\ &=0.0337\ \mathrm{mol} \end{aligned} \]

The apparent amount is approximately \(0.0337\ \mathrm{mol}\), compared with the actual dry gas amount of \(0.0331\ \mathrm{mol}\).

The student has overestimated the amount because the total pressure includes water vapour.

The positive y-intercept is the clue. It represents the constant water vapour pressure, not an increase in the amount of dry gas.

Question 4

A student investigates the pressure vs inverse-volume relationship using a gas syringe connected to a pressure sensor.

The connecting tube and sensor cavity have a combined volume of \(0.050\ \mathrm L\).

The syringe readings and absolute pressures are:

The gas is maintained at \(298.15\ \mathrm K\).

Calculate the amount of gas using the corrected total volume.

Explain why plotting against the inverse of the syringe reading alone would produce an incorrect result.

Solution 4

The corrected amount of gas is approximately \(0.0500\ \mathrm{mol}\). The connecting tube and sensor volume must be included because the gas occupies the entire connected space.

Step 1: Calculate the total volume.

Use:

\[ V_{\mathrm{total}}=V_{\mathrm{syringe}}+V_{\mathrm{dead}} \]

For each reading:

\[ \begin{aligned} V_1&=0.450+0.050=0.500\ \mathrm L\\ V_2&=0.750+0.050=0.800\ \mathrm L\\ V_3&=0.950+0.050=1.000\ \mathrm L \end{aligned} \]

Step 2: Calculate the inverse total volumes.

\[ \begin{aligned} \frac{1}{0.500}&=2.00\ \mathrm{L^{-1}}\\ \frac{1}{0.800}&=1.25\ \mathrm{L^{-1}}\\ \frac{1}{1.000}&=1.00\ \mathrm{L^{-1}} \end{aligned} \]

Step 3: Calculate the gradient.

Using the first and last readings:

\[ \begin{aligned} m&=\frac{248-124}{2.00-1.00}\\ &=124\ \mathrm{kPa\,L} \end{aligned} \]

Step 4: Calculate the amount of gas.

Since:

\[ m=nRT \]

Rearrange:

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

Substitute:

\[ \begin{aligned} n&=\frac{124}{8.314\times298.15}\\ &=\frac{124}{2478.82}\\ &=0.0500\ \mathrm{mol} \end{aligned} \]

Why is the correction important?

The tempting approach is to use the syringe scale reading as the total gas volume.

However, the gas also occupies the connecting tube and pressure sensor.

Using the syringe readings alone gives reciprocal volumes of approximately:

\[ 2.22,\quad1.33,\quad1.05\ \mathrm{L^{-1}} \]

These values do not produce the correct straight-line relationship.

Using the first and last readings would give an apparent gradient of approximately \(106\ \mathrm{kPa\,L}\), rather than \(124\ \mathrm{kPa\,L}\).

This would underestimate the calculated amount of gas.

The important lesson is that the inverse-volume axis must use the actual volume occupied by the gas, not simply the volume marked on the syringe.

Question 5

A variable-volume chamber contains an ideal gas. It is connected to an absolute-pressure sensor, a controllable outlet valve, and a heater.

Initially, the gas has:

• Amount: \(0.0500\ \mathrm{mol}\)
• Temperature: \(300\ \mathrm K\)
• Volume: \(1.00\ \mathrm L\)

During the experiment, the operator allows some gas to escape through the valve while also heating the remaining gas.

All measured points, including intermediate readings, produce a perfect straight line through the origin on a pressure vs inverse-volume graph.

The gradient is:

\[ m=124.7\ \mathrm{kPa\,L} \]

Independent measurements show that the final amount of gas is \(0.0400\ \mathrm{mol}\), and the final volume is \(0.500\ \mathrm L\).

Assume the ideal gas equation applies throughout.

a) Calculate the final temperature required for the final reading to lie on the original straight line.

b) Calculate the final absolute pressure.

c) Explain why the straight-line graph does not prove that Boyle’s law was followed.

d) Suppose the final temperature had instead been measured as \(360\ \mathrm K\). Calculate the resulting pressure and determine how its plotted point would compare with the original line.

Solution 5

The final temperature must be \(375\ \mathrm K\), and the final absolute pressure is approximately \(249.4\ \mathrm{kPa}\). The straight-line graph does not prove that the amount and temperature remained constant.

This question tests an important limitation of interpreting the gradient.

A straight line through the origin normally suggests that pressure is directly proportional to inverse volume.

However, the ideal gas equation tells us exactly what the gradient represents:

\[ m=nRT \]

It only measures the product of \(n\), \(R\), and \(T\).

Part a: Calculate the final temperature.

The initial amount and temperature are:

\[ n_1=0.0500\ \mathrm{mol} \]

\[ T_1=300\ \mathrm K \]

The initial value of the gradient is:

\[ \begin{aligned} m&=n_1RT_1\\ &=0.0500\times8.314\times300\\ &=124.71\ \mathrm{kPa\,L} \end{aligned} \]

This agrees with the measured gradient of \(124.7\ \mathrm{kPa\,L}\).

For the final state:

\[ m=n_2RT_2 \]

Rearrange:

\[ T_2=\frac{m}{n_2R} \]

Substitute:

\[ \begin{aligned} T_2&=\frac{124.71}{0.0400\times8.314}\\ &=\frac{124.71}{0.33256}\\ &=375\ \mathrm K \end{aligned} \]

The temperature must increase to \(375\ \mathrm K\).

Part b: Calculate the final absolute pressure.

Use the ideal gas equation:

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

Substitute:

\[ \begin{aligned} P_2&=\frac{0.0400\times8.314\times375}{0.500}\\ &=\frac{124.71}{0.500}\\ &=249.42\ \mathrm{kPa} \end{aligned} \]

Therefore:

\[ \boxed{P_2=249.4\ \mathrm{kPa}} \]

The pressure doubles compared with the initial pressure because the volume halves.

Part c: Does the graph prove Boyle’s law?

No.

Boyle’s law requires both the amount of gas and temperature to remain constant.

Neither condition has been maintained here:

• The amount decreases from \(0.0500\) to \(0.0400\ \mathrm{mol}\).
• The temperature increases from \(300\) to \(375\ \mathrm K\).

However, notice that:

\[ n_1T_1=0.0500\times300=15.0 \]

And:

\[ n_2T_2=0.0400\times375=15.0 \]

The product \(nT\) remains constant.

Since:

\[ m=nRT \]

and \(R\) is constant, the gradient remains unchanged.

This means the graph is indistinguishable from the graph of a sealed sample maintained at constant temperature.

The tempting conclusion is that a straight line through the origin proves Boyle’s law.

The actual conclusion is more limited: under the ideal gas assumption, the graph shows that \(nT\) remains constant. It does not independently prove that \(n\) and \(T\) are each constant.

To establish Boyle’s law experimentally, you must independently control or measure both conditions.

Part d: What if the final temperature is \(360\ \mathrm K\)?

Now the amount and temperature are:

\[ n=0.0400\ \mathrm{mol} \]

\[ T=360\ \mathrm K \]

Calculate the new pressure:

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

Substitute:

\[ \begin{aligned} P&=\frac{0.0400\times8.314\times360}{0.500}\\ &=\frac{119.72}{0.500}\\ &=239.4\ \mathrm{kPa} \end{aligned} \]

The pressure is now approximately \(239.4\ \mathrm{kPa}\).

Compare this with the pressure predicted by the original straight line:

\[ P_{\mathrm{line}}=\frac{124.7}{0.500} \]

\[ P_{\mathrm{line}}=249.4\ \mathrm{kPa} \]

The new pressure is approximately \(10.0\ \mathrm{kPa}\) lower, or 4% below the original line.

The reason is that the product \(nT\) has decreased.

The graph therefore bends away from the original relationship at this final point.

This is why interpreting a graph requires more than recognising its shape. You must understand which physical quantities determine that shape and whether the experiment actually controlled them.

The next useful step is learning when to move beyond Boyle’s law and use the combined gas law or the full ideal gas equation. If the amount of gas can also change, keep \(n\) in \(PV=nRT\) rather than assuming it is constant.