How to Read Pressure-Volume and Temperature Gas Graphs

You’ve drawn a straight line on a pressure-temperature graph and labelled it \(P\propto T\), but the temperature axis is in degrees Celsius. Or you’ve drawn pressure increasing with volume because both variables are changing. These are common exam mistakes: choosing a graph shape before checking the experimental conditions.

Try this prediction. A sealed, rigid container of gas warms from \(20^\circ\mathrm C\) to \(40^\circ\mathrm C\). Does its pressure double because the Celsius temperature doubles?

No. The temperatures are approximately \(293\ \mathrm K\) and \(313\ \mathrm K\), so the pressure increases by a factor of \(313/293\approx1.07\), assuming ideal gas behaviour. That’s about a 7% increase, not a 100% increase.

The fix is to identify what stays constant, choose the correct relationship, and only then decide what the graph should look like.

01First, check what must stay constant

The starting point is the ideal gas equation:

\[
PV=nRT
\]

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

For a fixed amount of gas, \(n\) stays constant. If the gas behaves ideally, the equation tells you how pressure, volume, and temperature are related.

The critical detail is that each simple gas law requires a particular variable to stay constant.

These are the three relationships to recognise. The graph shapes aren’t arbitrary pictures to memorise; they follow from the conditions and the ideal gas equation.

01Pressure against volume: inverse, not direct

Imagine compressing a sealed syringe slowly while keeping the gas at the same temperature. The same amount of gas must fit into a smaller space.

Think of the gas particles as guests in a room. If the room shrinks but everyone stays, the guests reach the walls more frequently. In the gas model, the particles collide with the container walls more frequently per unit area, increasing the pressure.

The analogy has a limit: molecules don’t make decisions about personal space, and the ideal gas model simplifies their size and intermolecular attractions. The useful idea is simply that reducing the volume increases the pressure when temperature and amount of gas stay fixed.

This is Boyle’s law. For a fixed amount of gas at constant temperature,

\[
PV=k
\]

where \(k\) is a constant for that sample under those conditions. Rearranging gives

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

Pressure is therefore inversely proportional to volume.

• Double the volume, and the pressure halves.
• Halve the volume, and the pressure doubles.
• Increase the volume by a factor of three, and the pressure becomes one-third of its original value.

The graph of \(P\) against \(V\) is a decreasing curve, not a straight line. A graph of \(P\) against \(1/V\), however, would be a straight line through the origin.

02Pressure against temperature: the volume must stay fixed

Now consider a sealed, rigid container of gas. Heat it while keeping the container’s volume unchanged.

The particles move faster as the temperature rises. They collide with the walls more forcefully and, on average, more frequently. The pressure increases.

Because volume and amount of gas are constant, the ideal gas equation gives

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

This is the pressure law, also commonly called Gay-Lussac’s law.

The useful equation is

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

The temperature must be in kelvin.

Why a Celsius graph can trick you

A pressure-temperature graph can be a straight line without showing direct proportionality.

A directly proportional relationship has the form \(y=kx\). Its graph is a straight line through the origin. A straight line with a non-zero intercept is linear, but it isn’t directly proportional.

Kelvin is essential because its zero corresponds to absolute zero. Celsius has a different zero point:

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

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

At constant volume, pressure is proportional to \(t+273.15\), not to \(t\) alone. At \(0^\circ\mathrm C\), a gas still has a positive absolute pressure.

The extrapolated line reaches zero pressure at approximately \(-273.15^\circ\mathrm C\), equivalent to \(0\ \mathrm K\), in the ideal gas model. This is a mathematical extrapolation, not a claim that a real gas can be heated or cooled through every point on that line while remaining a gas.

The exam rule is simple: a pressure-temperature graph is directly proportional only when temperature is expressed in kelvin and pressure is absolute, under the stated ideal-gas conditions.

03Volume against temperature: the pressure must stay fixed

Consider a gas trapped beneath a movable piston. If the piston can rise freely against a constant external load, the gas can expand as it is heated while its pressure remains approximately constant.

The amount of gas stays fixed, the pressure stays fixed, and the volume increases with temperature.

From the ideal gas equation,

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

This is Charles’s law. Its equation is

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

Again, temperature must be in kelvin.

The volume-temperature graph is a straight line through the origin when the horizontal axis is in kelvin. If the axis is in Celsius, the graph remains a straight line, but it does not pass through the origin.

The extrapolated zero-volume point is not a physically attainable state for an ordinary gas. Real gases change phase, and the ideal gas model is not valid all the way to absolute zero.

Notice the pattern across the three laws:

• For \(P\) against \(V\), temperature must stay constant, and the relationship is inverse.
• For \(P\) against \(T\), volume must stay constant, and the relationship is direct in kelvin.
• For \(V\) against \(T\), pressure must stay constant, and the relationship is direct in kelvin.

A memorised graph shape won’t save you if the experimental condition changes.

05A decision method you can reuse in an exam

When you see a gas graph, work through these steps in order.

Step 1: Read both axes.

Identify the variables and their units. Check whether temperature is in kelvin or Celsius, and whether the horizontal axis is volume or reciprocal volume.

Step 2: Identify what stays constant.

Look for wording such as “at constant temperature”, “in a rigid container”, or “at constant pressure”. Also check that the amount of gas is fixed.

If the required condition isn’t satisfied, you cannot automatically apply the corresponding simple gas law.

Step 3: Write the relationship before judging the graph.

Choose the appropriate expression:

\[
\begin{aligned}
T,n\text{ constant:}\quad &PV=k\\
V,n\text{ constant:}\quad &\frac{P}{T}=k\\
P,n\text{ constant:}\quad &\frac{V}{T}=k
\end{aligned}
\]

Here, \(k\) is a constant for the specified conditions.

Step 4: Check the graph’s shape and intercept.

• An inverse relationship between \(P\) and \(V\) should produce a decreasing curve.
• A direct relationship between \(P\) and \(T\) in kelvin should produce a straight line through the origin.
• A direct relationship between \(V\) and \(T\) in kelvin should also produce a straight line through the origin.

Remember that straight-line graphs against Celsius temperature can have non-zero intercepts.

Step 5: Test the actual data.

Don’t rely only on how a graph looks. Calculate the relevant ratio or product for multiple points. If the stated conditions are exact, the values should satisfy the relationship.

If they don’t, check the axes, units, pressure reference, measurement errors, and whether the claimed constant condition was actually maintained.

06Questions and solutions

Question 1

A fixed amount of ideal gas has a pressure of \(90\ \mathrm{kPa}\) and a volume of \(480\ \mathrm{cm^3}\). Its temperature remains constant while its volume decreases to \(360\ \mathrm{cm^3}\).

Calculate the final pressure and state whether the pressure-volume graph should rise or fall as volume increases.

Solution 1

The final pressure is \(120\ \mathrm{kPa}\), and the pressure-volume graph must fall as volume increases.

At constant temperature, Boyle’s law applies:

\[
P_1V_1=P_2V_2
\]

Rearranging and substituting,

\[
\begin{aligned}
P_2&=\frac{P_1V_1}{V_2}\\[4pt]
&=\frac{(90\ \mathrm{kPa})(480\ \mathrm{cm^3})}
{360\ \mathrm{cm^3}}\\[4pt]
&=120\ \mathrm{kPa}
\end{aligned}
\]

The volume falls to three-quarters of its initial value, so the pressure rises to four-thirds of its initial value. The graph is an inverse curve, not a rising straight line.

Question 2

A fixed amount of ideal gas expands at constant pressure. Its volume is \(250\ \mathrm{cm^3}\) at \(20^\circ\mathrm C\) and \(300\ \mathrm{cm^3}\) at \(80^\circ\mathrm C\). The volumes have been rounded to the nearest \(5\ \mathrm{cm^3}\).

A student claims that these readings are inconsistent with Charles’s law because the ratio \(V/t\), using Celsius temperature \(t\), is not constant.

Is the student correct? Explain using the appropriate ratios, and state what the graph should look like if the horizontal axis is in Celsius.

Solution 2

The student is incorrect: the readings are consistent with Charles’s law to the stated rounding, but the Celsius temperature must not be used in the proportionality ratio.

Convert the temperatures:

\[
\begin{aligned}
T_1&=20+273.15=293.15\ \mathrm K\\
T_2&=80+273.15=353.15\ \mathrm K
\end{aligned}
\]

Now compare volume per kelvin:

\[
\frac{V_1}{T_1}
=\frac{250}{293.15}
\approx0.853\ \mathrm{cm^3\,K^{-1}}
\]

\[
\frac{V_2}{T_2}
=\frac{300}{353.15}
\approx0.850\ \mathrm{cm^3\,K^{-1}}
\]

The ratios are close enough to be consistent with the rounded readings. The small difference does not establish a failure of the law.

The graph of volume against Celsius temperature is still a straight line, but it does not pass through the origin. A straight line is not automatically a directly proportional relationship; for proportionality, the graph must pass through the origin when the variables are expressed in the appropriate form.

Question 3

A fixed amount of ideal gas is held at constant volume. A student plots absolute pressure \(P\), in kilopascals, against temperature \(T\), in kelvin. The graph is a straight line described by

\[
P=0.40T+20
\]

The student argues that the graph must be correct because the pressure increases linearly with temperature.

Is this conclusion justified? Identify the problem with the graph and give one plausible experimental explanation.

Solution 3

No. The graph is straight, but it does not show the required direct proportionality for an ideal gas at constant volume.

The pressure law requires

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

Therefore, the graph of absolute pressure against kelvin temperature should pass through the origin.

The equation \(P=0.40T+20\) has a positive pressure intercept of \(20\ \mathrm{kPa}\) at \(0\ \mathrm K\). That contradicts the ideal-gas relationship under the stated conditions.

A plausible experimental explanation is a pressure sensor with a zero offset. Alternatively, the volume or amount of gas may not actually have remained constant. Because the question specifies an ideal gas and constant conditions, the graph cannot be accepted merely because it is linear.

Question 4

A fixed amount of ideal gas initially has a pressure of \(100\ \mathrm{kPa}\), a volume of \(600\ \mathrm{cm^3}\), and a temperature of \(300\ \mathrm K\).

The gas is compressed to \(400\ \mathrm{cm^3}\) while its temperature rises to \(450\ \mathrm K\).

Calculate the final pressure. Explain why applying Boyle’s law alone would give an incorrect answer.

Solution 4

The final pressure is \(225\ \mathrm{kPa}\). Boyle’s law alone would give \(150\ \mathrm{kPa}\), but that result assumes the temperature remains constant.

Here, both volume and temperature change. The appropriate relationship for a fixed amount of ideal gas is the combined gas law:

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

Rearrange for the final pressure:

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

Substitute the values:

\[
\begin{aligned}
P_2&=\frac{(100\ \mathrm{kPa})(600\ \mathrm{cm^3})(450\ \mathrm K)}
{(300\ \mathrm K)(400\ \mathrm{cm^3})}\\[4pt]
&=100\times\frac{600}{400}\times\frac{450}{300}\\[4pt]
&=(100\ \mathrm{kPa})(1.5)(1.5)\\[4pt]
&=225\ \mathrm{kPa}
\end{aligned}
\]

The compression increases the pressure because the volume decreases. The heating also increases the pressure because the temperature rises. The two effects combine to produce a pressure \(2.25\) times the initial value.

Using Boyle’s law alone would ignore the temperature change and underestimate the final pressure.

Question 5

A fixed amount of ideal gas follows a three-stage process. All pressures are absolute.

The initial state, A, is:

\[
P_A=100\ \mathrm{kPa},\quad
V_A=2.0\ \mathrm L,\quad
T_A=300\ \mathrm K
\]

The gas then undergoes these changes:

• A to B: It is heated to \(450\ \mathrm K\) at constant volume.
• B to C: It expands at constant pressure until its volume is \(3.0\ \mathrm L\).
• C to D: It is cooled at constant pressure until its temperature returns to \(300\ \mathrm K\).

A student claims that state D must have a volume of \(2.0\ \mathrm L\), because the gas has returned to its initial temperature.

Determine the pressure, volume, and temperature at B, C, and D. Decide whether the student’s claim is correct, and explain why the tempting argument fails.

Solution 5

The student’s claim is incorrect. The actual final state is \(P_D=150\ \mathrm{kPa}\), \(V_D=1.33\ \mathrm L\), and \(T_D=300\ \mathrm K\), to three significant figures.

The key is to apply the correct gas law to each stage separately. Returning to the initial temperature does not guarantee a return to the initial volume if the pressure has changed.

Step 1: Find state B.

From A to B, volume and amount of gas are constant. The pressure law applies:

\[
\frac{P_A}{T_A}=\frac{P_B}{T_B}
\]

Therefore,

\[
\begin{aligned}
P_B&=P_A\frac{T_B}{T_A}\\[4pt]
&=(100\ \mathrm{kPa})\frac{450\ \mathrm K}{300\ \mathrm K}\\[4pt]
&=150\ \mathrm{kPa}
\end{aligned}
\]

The volume remains \(2.0\ \mathrm L\), so state B is

\[
P_B=150\ \mathrm{kPa},\quad
V_B=2.0\ \mathrm L,\quad
T_B=450\ \mathrm K
\]

Step 2: Find state C.

From B to C, pressure and amount of gas are constant. Charles’s law applies:

\[
\frac{V_B}{T_B}=\frac{V_C}{T_C}
\]

Rearrange for temperature:

\[
\begin{aligned}
T_C&=T_B\frac{V_C}{V_B}\\[4pt]
&=(450\ \mathrm K)\frac{3.0\ \mathrm L}{2.0\ \mathrm L}\\[4pt]
&=675\ \mathrm K
\end{aligned}
\]

The pressure remains \(150\ \mathrm{kPa}\), so state C is

\[
P_C=150\ \mathrm{kPa},\quad
V_C=3.0\ \mathrm L,\quad
T_C=675\ \mathrm K
\]

Notice that expanding the gas at constant pressure requires its temperature to rise. Volume and temperature cannot be assigned independently.

Step 3: Find state D.

From C to D, pressure remains \(150\ \mathrm{kPa}\), and the final temperature is \(300\ \mathrm K\). Again, use Charles’s law:

\[
\frac{V_C}{T_C}=\frac{V_D}{T_D}
\]

Therefore,

\[
\begin{aligned}
V_D&=V_C\frac{T_D}{T_C}\\[4pt]
&=(3.0\ \mathrm L)\frac{300\ \mathrm K}{675\ \mathrm K}\\[4pt]
&=1.33\ \mathrm L
\end{aligned}
\]

The actual final state is

\[
P_D=150\ \mathrm{kPa},\quad
V_D=1.33\ \mathrm L,\quad
T_D=300\ \mathrm K
\]

Step 4: Check why the student’s final state is impossible.

The student assumes that returning to \(300\ \mathrm K\) forces the volume to return to \(2.0\ \mathrm L\). That would only follow if the pressure also returned to its initial value.

For a fixed amount of ideal gas,

\[
\frac{PV}{T}=nR
\]

This quantity must remain constant throughout the process.

At the initial state,

\[
\frac{P_AV_A}{T_A}
=\frac{(100)(2.0)}{300}
=0.667\ \mathrm{kPa\,L\,K^{-1}}
\]

If the student’s claimed final state were correct, its value would be

\[
\frac{(150)(2.0)}{300}
=1.00\ \mathrm{kPa\,L\,K^{-1}}
\]

That is inconsistent with the same fixed amount of gas. Using the correct final volume instead gives

\[
\frac{(150)(1.33)}{300}
\approx0.667\ \mathrm{kPa\,L\,K^{-1}}
\]

The gas returns to its initial temperature but not its initial pressure. Its volume must therefore differ from its initial volume.

The deeper lesson is that every stage must obey its own constant-condition relationship, while all states must also remain consistent with the ideal gas equation. Three individually plausible graph segments do not make an impossible final state valid.

07The 20-second exam self-check

Before submitting a gas-graph answer, ask yourself:

• Have I identified the correct axes and units?
• Have I stated which variable and amount of gas stay constant?
• Have I converted Celsius temperatures to kelvin before using a ratio?
• Have I used \(PV=k\), \(P/T=k\), or \(V/T=k\) under the correct conditions?
• Have I checked whether the graph should be curved or straight, and whether a proportional graph should pass through the origin?
• If the data contradict the graph, have I checked the assumptions, pressure reference, units, and possible measurement errors?

The next useful step is the combined gas law. It lets you handle a change in pressure, volume, and temperature together, provided the amount of gas stays fixed:

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

Instead of forcing one simple gas law onto an entire process, identify which conditions hold at each stage-or use the combined gas law when several variables change at once.