Choosing R When Pressure Is in kPa and Volume Is in L

A gas sample has a pressure of \(120\ \mathrm{kPa}\), a volume of \(2.00\ \mathrm{L}\), and a temperature of \(27^\circ\mathrm{C}\). You want to use the ideal gas equation to calculate the amount of gas.

There is one decision to make first: which value of the gas constant, \(R\), should you use?

For pressure in kilopascals (kPa), volume in litres (L), temperature in kelvin (K), and amount in moles (mol), use:

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

You do not need to convert kilopascals into pascals or litres into cubic metres. The value \(8.314\) is already compatible with kPa and L.

However, there are a few traps worth understanding, particularly when experimental pressure readings are involved.

01Why is R equal to 8.314 when pressure is in kPa and volume is in L?

Start with the ideal gas equation:

\[ PV=nRT \]

Where:

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

The gas constant is universal. Its numerical value changes depending on the units used, but the physical constant itself does not change.

Here are the common combinations:

The corresponding units must also be included.

For example, when pressure is in kPa and volume is in L:

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

When pressure is in Pa and volume is in cubic metres:

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

Both expressions represent the same physical constant.

Why do both combinations use 8.314?

The important relationship is:

\[ 1\ \mathrm{kPa}=1000\ \mathrm{Pa} \]

And:

\[ 1\ \mathrm{L}=0.001\ \mathrm{m^3} \]

Multiply these conversions:

\[ \begin{aligned} 1\ \mathrm{kPa\,L} &=1000\ \mathrm{Pa}\times0.001\ \mathrm{m^3}\\ &=1\ \mathrm{Pa\,m^3}\\ &=1\ \mathrm{J} \end{aligned} \]

This means that a pressure-volume product expressed in kPa L has the same numerical value as the corresponding energy expressed in joules.

That is why:

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

is numerically identical to:

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

The practical lesson is simple: choose \(R\) based on the combination of pressure and volume units, not just one of them.

02Three things to check before substituting into PV = nRT

Choosing the correct gas constant solves one problem. You still need to check the other quantities.

Temperature must be in kelvin

The ideal gas equation requires absolute temperature, not degrees Celsius.

Convert using:

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

For example:

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

Do not substitute \(25\) directly into the equation.

This is particularly important because the relationship between pressure, volume, and temperature depends on absolute temperature.

Pressure must be absolute pressure

This is an easy detail to miss in experimental questions.

There are two common ways of reporting pressure:

• Absolute pressure: pressure measured relative to a perfect vacuum.
• Gauge pressure: pressure measured relative to the surrounding atmospheric pressure.

The ideal gas equation requires absolute pressure.

If you are given gauge pressure, convert it using:

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

For example, suppose a pressure sensor reads \(20.0\ \mathrm{kPa}\) gauge pressure, while atmospheric pressure is \(101.3\ \mathrm{kPa}\).

Then:

\[ P_{\mathrm{absolute}}=20.0+101.3 \]

\[ P_{\mathrm{absolute}}=121.3\ \mathrm{kPa} \]

You must use \(121.3\ \mathrm{kPa}\) in the ideal gas equation.

Volume must match the chosen value of R

If your volume is already in litres, you can use it directly with \(R=8.314\).

However, be careful with other units.

Your amount of gas must also be in moles.

Your exam decision rule

Before using the ideal gas equation, check these four things:

If all four conditions are satisfied, use:

\[ \boxed{PV=n(8.314)T} \]

with the appropriate units.

03How an experiment can determine R

The gas constant is not just a number to memorise. It can also be determined experimentally by measuring the pressure, volume, temperature, and amount of a gas sample.

Consider an experiment designed to test the ideal gas equation.

A known amount of dry gas is sealed inside a rigid vessel of known volume. The vessel is placed in a temperature-controlled water bath.

A pressure sensor is connected to the vessel, and a thermometer measures the temperature of the bath. Once the gas reaches thermal equilibrium with the bath, the pressure and temperature can be recorded.

The experimental arrangement needs four known or measured quantities:

• The amount of gas, \(n\), in moles.
• The vessel volume, \(V\), in litres.
• The absolute temperature, \(T\), in kelvin.
• The absolute pressure, \(P\), in kilopascals.

Rearrange the ideal gas equation to make \(R\) the subject:

\[ R=\frac{PV}{nT} \]

If the gas behaves ideally, repeated measurements should give approximately the same value of \(R\), provided the measurements and corrections are reliable.

What should the experiment show?

Suppose the experiment uses:

• A rigid vessel with a volume of \(1.000\ \mathrm{L}\).
• A fixed amount of gas, \(0.0500\ \mathrm{mol}\).

The temperature is changed while the volume and amount of gas remain constant.

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

Using the first measurement:

\[ \begin{aligned} R&=\frac{PV}{nT}\\ &=\frac{116.4\times1.000}{0.0500\times280.0}\\ &=\frac{116.4}{14.0}\\ &=8.314 \end{aligned} \]

The units are:

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

Repeating the calculation for the other readings gives approximately \(8.317\) and \(8.313\).

These values are close to the accepted ideal gas constant of \(8.314\).

The small differences are consistent with rounding and experimental uncertainty.

The result supports the ideal gas relationship under the conditions of this experiment. It does not prove that every real gas behaves perfectly ideally.

What could make the experimental result inaccurate?

The original arrangement assumes that the gas is dry, that its amount is known, and that the pressure reading is absolute.

If any of these conditions is incorrect, the calculated value of \(R\) may be inaccurate.

A revised experimental design needs to address the following problems:

Notice something important: these are corrections to the measurements, not changes to the gas constant.

That distinction becomes particularly useful when working with experimental questions.

06The common mistake: Using 0.08206 because volume is in litres

This is one of the easiest ways to lose marks in an ideal gas calculation.

Suppose you have:

\[ P=100\ \mathrm{kPa} \]

\[ V=2.00\ \mathrm L \]

\[ T=300\ \mathrm K \]

You want to calculate the amount of gas.

A student remembers that the gas constant can be \(0.08206\) and chooses it because the volume is measured in litres.

That choice is incomplete.

The value:

\[ R=0.08206\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}} \]

requires pressure in atmospheres, not kilopascals.

Using \(8.314\) with kPa and L:

\[ \begin{aligned} n&=\frac{PV}{RT}\\ &=\frac{100\times2.00}{8.314\times300}\\ &=0.0802\ \mathrm{mol} \end{aligned} \]

Now compare that with using \(0.08206\) without converting the pressure:

\[ n=\frac{100\times2.00}{0.08206\times300} \]

\[ n=8.12\ \mathrm{mol} \]

The second result is approximately 101 times larger.

Why?

Because:

\[ 1\ \mathrm{atm}=101.325\ \mathrm{kPa} \]

If you want to use \(R=0.08206\), you must first convert the pressure:

\[ 100\ \mathrm{kPa}=0.987\ \mathrm{atm} \]

Using the converted pressure gives approximately the same answer:

\[ n=\frac{0.987\times2.00}{0.08206\times300} \]

\[ n=0.0801\ \mathrm{mol} \]

The small difference comes from rounding.

Remember: litres alone do not determine which value of \(R\) to use. Pressure and volume must be considered together.

07Ideal gas assumptions versus experimental corrections

There is one final distinction worth understanding.

The ideal gas equation is a model. It describes gas behaviour using simplifying assumptions.

In the ideal gas model:

• Gas particles have negligible individual volume compared with the container.
• There are no significant intermolecular attractions or repulsions between particles.
• Collisions are elastic.
• Particles move randomly.

Real gases do not satisfy these assumptions perfectly.

At sufficiently high pressures or low temperatures, the volume of the particles and intermolecular attractions become more significant.

The gas may then deviate from ideal behaviour.

However, this does not mean that the universal gas constant has changed.

It means that the ideal gas model is no longer an exact description of the gas.

How can you distinguish the two problems?

An optional way to describe non-ideal behaviour is through the compressibility factor, \(Z\):

\[ Z=\frac{PV}{nRT} \]

For an ideal gas:

\[ Z=1 \]

For a real gas, \(Z\) may differ from 1.

This is a useful extension because it separates two different problems:

• An incorrect calculated value of \(R\) caused by measurement or unit errors.
• A genuine deviation from ideal gas behaviour.

You should not change the value of \(R\) to compensate for an experimental error or a non-ideal gas.

08Questions and solutions

Question 1

A sealed container holds an ideal gas at an absolute pressure of \(105\ \mathrm{kPa}\), a volume of \(1.80\ \mathrm L\), and a temperature of \(25.0^\circ\mathrm C\).

Calculate the amount of gas in moles.

Solution 1

The amount of gas is approximately \(0.0762\ \mathrm{mol}\).

Since pressure is in kPa and volume is in L, use:

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

First, convert the temperature:

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

Rearrange the ideal gas equation:

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

Substitute:

\[ \begin{aligned} n&=\frac{105\times1.80}{8.314\times298.15}\\ &=\frac{189}{2478.83}\\ &=0.0762\ \mathrm{mol} \end{aligned} \]

The important point is that the pressure and volume units already match the chosen value of \(R\). Only the temperature required conversion.

Question 2

A chemistry student performs an experiment using a gas sample with the following measurements:

• Pressure: \(84.0\ \mathrm{kPa}\).
• Volume: \(375\ \mathrm{mL}\).
• Temperature: \(42.0^\circ\mathrm C\).
• Amount of gas: \(0.0120\ \mathrm{mol}\).

The student wants to calculate an experimental value of \(R\).

(a) Calculate \(R\) using pressure in kPa and volume in L.

(b) Calculate the numerical value of \(R\) if the student keeps the volume in mL instead.

(c) Explain why the numerical values in parts (a) and (b) are different.

Solution 2

The experimental values are approximately \(8.33\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\) and \(8330\ \mathrm{kPa\,mL\,mol^{-1}\,K^{-1}}\), respectively.

Part (a)

First, convert the volume:

\[ V=375\ \mathrm{mL}=0.375\ \mathrm L \]

Convert the temperature:

\[ T=42.0+273.15=315.15\ \mathrm K \]

Rearrange:

\[ R=\frac{PV}{nT} \]

Substitute:

\[ \begin{aligned} R&=\frac{84.0\times0.375}{0.0120\times315.15}\\ &=\frac{31.5}{3.7818}\\ &=8.33 \end{aligned} \]

Therefore:

\[ R=8.33\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}} \]

Part (b)

Keep the original volume in millilitres:

\[ R=\frac{84.0\times375}{0.0120\times315.15} \]

\[ R=\frac{31500}{3.7818} \]

\[ R\approx8329 \]

Therefore:

\[ R\approx8330\ \mathrm{kPa\,mL\,mol^{-1}\,K^{-1}} \]

Part (c)

The numerical values differ because:

\[ 1\ \mathrm L=1000\ \mathrm{mL} \]

Changing the volume unit from L to mL makes the numerical value of \(R\) 1000 times larger.

The physical gas constant has not changed. Only its numerical representation and units have changed.

Question 3

A student uses a pressure sensor to investigate a dry gas sample.

The following measurements are recorded:

• Gauge pressure: \(14.2\ \mathrm{kPa}\).
• Atmospheric pressure: \(98.7\ \mathrm{kPa}\).
• Volume: \(0.640\ \mathrm L\).
• Temperature: \(18.0^\circ\mathrm C\).
• Amount of gas: \(0.0299\ \mathrm{mol}\).

(a) Calculate the experimental value of \(R\).

(b) Calculate the value the student would obtain if they incorrectly used gauge pressure directly.

(c) Explain why the two results are so different.

Solution 3

The correct experimental value is approximately \(8.30\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\).

Part (a)

First, convert gauge pressure into absolute pressure:

\[ P_{\mathrm{absolute}}=14.2+98.7 \]

\[ P_{\mathrm{absolute}}=112.9\ \mathrm{kPa} \]

Convert temperature:

\[ T=18.0+273.15 \]

\[ T=291.15\ \mathrm K \]

Calculate \(R\):

\[ R=\frac{PV}{nT} \]

\[ \begin{aligned} R&=\frac{112.9\times0.640}{0.0299\times291.15}\\ &=\frac{72.256}{8.7054}\\ &=8.30 \end{aligned} \]

Part (b)

If gauge pressure is incorrectly substituted:

\[ R=\frac{14.2\times0.640}{0.0299\times291.15} \]

\[ R=\frac{9.088}{8.7054} \]

\[ R\approx1.04 \]

Part (c)

Gauge pressure measures pressure relative to atmospheric pressure, not relative to a vacuum.

The ideal gas equation requires absolute pressure.

The missing atmospheric pressure contribution is substantial in this experiment. Using gauge pressure directly therefore produces a severely underestimated value of \(R\).

The pressure reference must be corrected before using the gas equation.

Question 4

A chemistry student investigates a gas at relatively high pressure and obtains the following experimental data:

• Absolute pressure: \(1200\ \mathrm{kPa}\).
• Volume: \(0.100\ \mathrm L\).
• Amount of gas: \(0.0500\ \mathrm{mol}\).
• Temperature: \(300\ \mathrm K\).

The student checks the apparatus and confirms that the gas is dry, the pressure sensor is calibrated, and the readings are reproducible.

(a) Calculate the experimental value of \(R\).

(b) Calculate the compressibility factor, \(Z\), using the accepted ideal gas constant.

(c) The student concludes that the gas has its own gas constant of \(8.00\). Explain why this conclusion is incorrect.

(d) Identify two assumptions of the ideal gas model that may become unreliable under these conditions.

Solution 4

The experimental value is \(8.00\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\), and the compressibility factor is approximately \(0.962\).

Part (a)

Rearrange the ideal gas equation:

\[ R=\frac{PV}{nT} \]

Substitute:

\[ \begin{aligned} R&=\frac{1200\times0.100}{0.0500\times300}\\ &=\frac{120}{15}\\ &=8.00 \end{aligned} \]

Part (b)

The compressibility factor is:

\[ Z=\frac{PV}{nRT} \]

Using the accepted value of \(R\):

\[ \begin{aligned} Z&=\frac{1200\times0.100} {0.0500\times8.314\times300}\\ &=\frac{120}{124.71}\\ &=0.962 \end{aligned} \]

For an ideal gas, \(Z=1\).

The experimental result is therefore consistent with a deviation from ideal gas behaviour.

Part (c)

The student is incorrect because the universal gas constant does not depend on the identity of the gas.

The value \(8.00\) is an experimental ratio calculated using the ideal gas equation. It does not mean that the universal gas constant has changed.

Since the apparatus has been checked, the difference suggests that the ideal gas model does not describe the sample perfectly under these conditions.

Part (d)

Two ideal gas assumptions that may become unreliable are:

• Gas particles have negligible volume compared with the container.
• Intermolecular attractions and repulsions are negligible.

At high pressure, the particles are closer together. Their individual volumes and interactions become more significant.

The real gas can therefore depart from the ideal gas relationship even when the experimental measurements are accurate.

Question 5

This final problem combines pressure references, water vapour, and the assumptions behind the ideal gas equation.

A student investigates a fixed amount of dry gas inside a rigid vessel with a small pool of liquid water.

The vessel is sealed and placed in a temperature-controlled bath. The gas above the water is saturated with water vapour at each temperature.

A pressure sensor measures gauge pressure relative to the atmosphere.

The following illustrative measurements are recorded:

The vessel has a constant volume of \(0.750\ \mathrm L\). The amount of dry gas remains constant at \(0.0305\ \mathrm{mol}\).

Assume the dry gas behaves ideally, and use the supplied water vapour pressures.

(a) Calculate the absolute total pressure of the gas mixture in each trial.

(b) Calculate the pressure of the dry gas in each trial.

(c) Calculate the experimental value of \(R\) for each trial.

(d) A student argues that the results disprove the ideal gas equation because the gauge pressure increases by a factor of 5.4 when the temperature increases. Explain precisely why this reasoning is incorrect.

Solution 5

Both trials give approximately the same experimental gas constant:

\[ R\approx8.32\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}} \]

The results are consistent with ideal behaviour for the dry gas, within the precision of the supplied data.

The key is to recognise that the pressure sensor measures gauge pressure, while the gas mixture also contains water vapour.

Part (a)

First, convert gauge pressure into absolute pressure.

The relationship is:

\[ P_{\mathrm{total}}=P_{\mathrm{gauge}}+P_{\mathrm{atmospheric}} \]

For Trial 1:

\[ P_{\mathrm{total,1}}=4.0+100.0 \]

\[ P_{\mathrm{total,1}}=104.0\ \mathrm{kPa} \]

For Trial 2:

\[ P_{\mathrm{total,2}}=21.6+100.0 \]

\[ P_{\mathrm{total,2}}=121.6\ \mathrm{kPa} \]

These are the absolute pressures of the complete gas mixture, including water vapour.

Part (b)

We now need the pressure contributed by the dry gas alone.

The total pressure is the sum of the partial pressures of the dry gas and water vapour.

Therefore:

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

For Trial 1:

\[ P_{\mathrm{dry,1}}=104.0-3.17 \]

\[ P_{\mathrm{dry,1}}=100.83\ \mathrm{kPa} \]

For Trial 2:

\[ P_{\mathrm{dry,2}}=121.6-12.3 \]

\[ P_{\mathrm{dry,2}}=109.3\ \mathrm{kPa} \]

Notice that the water vapour pressure increases substantially when the temperature increases.

This happens because the liquid water produces a greater equilibrium vapour pressure at the higher temperature.

The amount of water vapour in the gas mixture therefore changes, even though the amount of dry gas stays constant.

Part (c)

Now calculate \(R\) for each trial.

First, convert the temperatures into kelvin.

Trial 1:

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

Trial 2:

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

Use:

\[ R=\frac{PV}{nT} \]

For Trial 1:

\[ \begin{aligned} R_1&=\frac{100.83\times0.750} {0.0305\times298.15}\\ &=\frac{75.6225}{9.0936}\\ &=8.316 \end{aligned} \]

For Trial 2:

\[ \begin{aligned} R_2&=\frac{109.3\times0.750} {0.0305\times323.15}\\ &=\frac{81.975}{9.8561}\\ &=8.317 \end{aligned} \]

Therefore, the two experimental estimates are:

\[ R_1=8.316 \]

\[ R_2=8.317 \]

Both are very close to the accepted value of \(8.314\).

The results are consistent with the same ideal gas constant at both temperatures.

Part (d)

The student’s conclusion is incorrect because they have compared gauge pressures rather than absolute dry-gas pressures.

The gauge pressure increases from \(4.0\ \mathrm{kPa}\) to \(21.6\ \mathrm{kPa}\), a factor of:

\[ \frac{21.6}{4.0}=5.4 \]

However, this ratio has no direct proportionality relationship with absolute temperature.

There are two reasons.

First, gauge pressure is measured relative to atmospheric pressure. It is not the pressure that belongs directly in the ideal gas equation.

Second, the total pressure includes water vapour. The water vapour pressure changes with temperature, so the total number of gas particles in the vessel is not constant.

The dry gas, however, has a constant amount and occupies a constant volume.

For that dry gas, the ideal gas equation predicts:

\[ \frac{P_{\mathrm{dry}}}{T}=\text{constant} \]

Compare the corrected pressure ratio:

\[ \frac{109.3}{100.83}\approx1.084 \]

With the absolute temperature ratio:

\[ \frac{323.15}{298.15}\approx1.084 \]

The ratios agree closely.

This is the important conceptual distinction: the pressure law applies to the dry gas’s partial pressure here, not to the uncorrected gauge pressure or the total pressure of the changing gas mixture.

A student who simply substitutes the sensor readings into \(PV=nRT\) misses both corrections and may incorrectly conclude that the gas constant changes with temperature.

09Where to go next

Once you can choose \(R\) correctly, the next step is recognising how pressure, volume, and temperature change together when one of these quantities is held constant.

Understanding those conditions makes it much easier to move between the ideal gas equation and the individual gas laws.

For further practice, see How to Read Pressure-Volume and Temperature Gas Graphs.

The key decision to carry into your next calculation is straightforward: match the units, use absolute pressure, convert temperature to kelvin, and distinguish measurement corrections from genuine deviations from ideal gas behaviour.