“Do we have to convert the Celsius to Kelvin?” is the question that I probably heard the most common when solving practice problems on gas laws in class.

And the short answer is yes, it is almost always necessary to convert the units of temperature to Kelvin. The reason for this is the units of the gas constant R. Although it may have different units depending on the units of the pressure and volume, in chemistry, it is most often going to be

\[R\;{\rm{ = }}\;{\rm{0}}{\rm{.08206}}\;\frac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}}\]

For example, a sample of hydrogen gas is added in a 5.80 L container at 56.0 °C. How many moles of the gas are present in the container if the pressure is 6.70 atm?

Let’s first write down what we are given:

V = 5.80 L

T = 56.0 ^oC

P = 6.70 atm

n(H₂) = ?

T = 273 + 56.0 = 329 K

To find the moles of hydrogen, we need the ideal gas law equation:

PV = nRT

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

\[{\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{6}}{\rm{.70}}\;\cancel{{{\rm{atm}}}}\; \times \;{\rm{5}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{ }} \times \;{\rm{329}}\;\cancel{{\rm{K}}}}}\; = \;1.44\;{\rm{mol}}\]

Notice how having the units in K allowed us to cancel them out.

The take-home message is whenever you are using a formula with R = 0.08206 L · atm / K · mol, change the units of temperature to Kelvin.

What Happens if I don’t Convert to Kelvin?

Let’s say you are working on a problem that does not use the ideal gas law equation.

For example, a gas sample is stored in a 429 mL container at 9.50°C and 2.20 atm. Calculate the pressure of the gas if the volume changes to 134 mL and the container is heated to 134.5 °C? Assume a constant amount of gas.

First, remember what we discuss about choosing the correct gas law. Instead of trying to remember which one to use, go with the combined gas law equation.

Write it down and remove the quantities that are not mentioned in the problem.

In this case, we have:

V₁ = 429 mL

V₂ = 134 mL

P₁ = 2.20 atm

T₁ = 9.50 ^oC

T₂ = 134.5 ^oC

P₂ = ?

Remember, the temperature must always be converted to K in gas problems, and we will see why in a second.

T₁ = 9.50 + 273 = 282.5 K

T₂ = 134.5 +273 = 407.5 K

Next, write the combined gas law equating and get rid of the constants. In this case, only the moles of the gas are constant:

\[\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}{{\rm{T}}_{\rm{2}}}}}\]

\[\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{V}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{V}}_{\rm{2}}}}}\; = \;\frac{{{\rm{2}}{\rm{.20}}\;{\rm{atm}}\;{\rm{ \times }}\;{\rm{429}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;{\rm{407}}{\rm{.5}}\;\cancel{{\rm{K}}}}}{{{\rm{282}}{\rm{.5}}\;\cancel{{{\rm{K}}\;}}\;{\rm{134}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{10}}{\rm{.2}}\;{\rm{atm}}\]

Notice that we did not need to convert the ml to L because they cancel out in the equation. The volume must be in L though when the R, expressed using L, is part of the calculation.

So, let’s say we did not convert the units of T to Kelvin. The answer would have been:

                                                            \[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}{{\rm{V}}_{\rm{2}}}}}\; = \;\frac{{{\rm{2}}.{\rm{20}}\;{\rm{atm}}\; \times \;{\rm{429}}\;\cancel{{{\rm{mL}}}}\; \times \;{\rm{134}}.{\rm{5}}\;\cancel{{\rm{K}}}}}{{{\rm{9}}.{\rm{5}}\;\cancel{{{\rm{K}}\;}}\;{\rm{134}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;99.7\;{\rm{atm}}\]

We are getting a wrong answer when the units of temperature are not in Kelvin.

The difference between volume and temperature is that for the volume, we multiply or divide by 1000 when converting between ml and L, while for temperature, we add 273 which does not change both numbers with an equal magnitude.

To illustrate this, consider a simple example. Six divided by 2 is 3, and if we multiply both parts of the fraction by 20, we are going to get the same answer. However, if we add 20 to each part of the fraction, like we do for temperature, the answer is going to be different.

To summarize, always convert the temperature to Kelvin when using a gas law problem. The exceptions would be when R is different than 0.08206 L · atm / K · mol.

How are the Units and Value of R Obtained?

Let’s for a moment go back to how we derive the ideal gas law equation to see where the units of R are coming from. The ideal gas law is derived from the individual gas laws such as the Boyle’s law, Charles’ law, etc. Remember, when discussing those, we used a pump with a freely moving plunger and by changing the parameters one by one, we were able to get the following correlations:

We can combine all of these to get an expression showing how, for example, the volume of an ideal gas is correlated to the temperature, pressure, and the moles of the gas:

To bring in the equal sign, we then introduce a constant:

This can be explained using the example of a car dealership income. The income depends on the number of sales which we can represent as:

                                             income ~ number of cars sold 

However, we cannot say income = number of cars sold, so to switch an equal sign, we need to introduce a constant. This can be the price of the car transforming the equation to:

Income  = price x number of cars

Now, for the parameters of ideal gases, the constant is the R, which is the ideal gas constant that has the same value for all gases because it is independent of the identity of the gas:

\[V\;{\rm{ = }}\,R\, \times \,\frac{{nT}}{P}\]

And finally, by rearranging it, we obtain the ideal gas law equation:

Now, we can also rearrange the ideal gas law equation to get an expression for the R:

\[R\;{\rm{ = }}\,\frac{{PV}}{{nT}}\]

Conventionally, the units for pressure are atm, volume is given in liters, the SI unit of temperature is Kelvin, and for the amount of chemicals is moles. Therefore, the units of R would be:

\[R\;{\rm{ = }}\,\frac{{PV}}{{nT}}\; = \,\frac{{{\rm{atm}}\, \cdot \,{\rm{L}}}}{{{\rm{mol}}\, \cdot \,{\rm{K}}}}\]

How do we obtain the numeric value of R?

Theoretically, we could use any quantity of any ideal gas, measure its volume and pressure, and plug in these numbers to obtain the value of R. However, to keep a standard and consistency, this is done for a gas at standard temperature and pressure, often abbreviated STP. This is when we have one mole of an ideal gas at 0 ^oC (273.15 K) and 1 atm pressure. It is shown experimentally that under these conditions, 1 mole of an ideal gas occupies 22.414 L, which is called the molar volume of the gas.

Using these numbers, a numeric value and units of R can be obtained:

\[R\;{\rm{ = }}\,\frac{{PV}}{{nT}}\; = \,\frac{{{\rm{(1}}\,{\rm{atm)}}\,{\rm{(22}}{\rm{.414 L)}}}}{{{\rm{(1}}\,{\rm{mol)}}\,{\rm{(273}}{\rm{.15}}\,{\rm{K)}}}}\]

\[R\;{\rm{ = }}\,{\rm{0}}{\rm{.08206}}\frac{{{\rm{atm}}\, \cdot \,{\rm{L}}}}{{{\rm{mol}}\, \cdot \,{\rm{K}}}}\]

As you may wonder, this is not the only value of R because we can change for example the units of pressure. For example, when we use the SI units for pressure and volume as well (pascals and meters cubed respectively), R is obtained in joules per kelvin per mole: R = 8.314 J · K^-1 · mol^-1.

For most general chemistry classes, you are going to use the R = 0.08206 L · atm / K · mol.

And here are some summary gas law problems and if you are not sure how to start, make sure to check the articles on the gas law chapter.

Check Also

• Boyle’s Law
• Charle’s Law
• Gay-Lussac’s Law
• Avogadro’s Law
• The Ideal Gas Law
• Ideal-Gas Laws
• Combined Gas Law Equation
• How to Know Which Gas Law Equation to Use
• Molar Mass and Density of Gases
• Graham’s Law of Effusion and Diffusion
• Graham’s Law of Effusion Practice Problems
• Dalton’s Law of Partial Pressures
• Mole Fraction and Partial Pressure of the Gas
• Gases in Chemical Reactions
• Gases-Practice Problems