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Application of the Ideal Gas Law

Apply the ideal gas law to density, molar mass, and compression problems. Check units, absolute pressure, kelvin temperatures, and STP conventions.

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Application of the Ideal Gas Law

The ideal gas law connects pressure, volume, temperature, and amount of gas. It can also be used to estimate gas density or molar mass. Before substituting numbers, check the temperature scale, the pressure definition, and the units of the gas constant.

1. What the equation means

PV=nRT.PV=nRT.

Here, PP is absolute pressure, VV is volume, nn is the amount of gas in moles, TT is thermodynamic temperature in kelvin, and RR is the gas constant.

Convert Celsius temperatures using TK=t∘C+273.15\frac{T}{\mathrm{K}}=\frac{t}{{}^\circ\mathrm{C}}+273.15. Use 273273 only when that approximation is appropriate to the question.

Diagram linking pressure, volume, amount of gas, and temperature through the ideal gas law.

The illustration identifies the four variables; it does not specify measurement conditions. If a pressure gauge gives pressure relative to the atmosphere, convert it to absolute pressure before using the equation.

Real gases often approximate ideal behavior at low pressure and at temperatures well above condensation. The approximation becomes less reliable at high pressure or near condensation.

2. Choose a consistent set of units

Approximate value and units of RRPressureVolume
0.08206 L atm mol−1 K−10.08206\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}atmL
8.314 Pa m3 mol−1 K−18.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}Pam³
8.314 kPa L mol−1 K−18.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}kPaL

The last two rows have the same numerical value because 1 Pa m3=1 kPa L=1 J1\ \mathrm{Pa\,m^3}=1\ \mathrm{kPa\,L}=1\ \mathrm{J}. In every row, use moles for nn and kelvin for TT.

Do not combine pressure in Pa with volume in L and substitute R=8.314R=8.314 without converting one of those units.

3. Density and molar mass

Substitute n=m/Mn=m/M, where mm is mass and MM is molar mass:

PV=mMRT.PV=\frac{m}{M}RT.

Since density is ρ=m/V\rho=m/V,

PM=ρRT,M=ρRTP.PM=\rho RT,\qquad M=\frac{\rho RT}{P}.

For example, density in g/L with RR in L·atm·mol⁻¹·K⁻¹ and pressure in atm gives molar mass in g/mol.

Example 1: Estimating molar mass

A gas has density 2.62 g/L2.62\ \mathrm{g/L} at 27∘C27^\circ\mathrm{C} and 1.00 atm1.00\ \mathrm{atm}. Assuming ideal behavior, estimate its molar mass.

Use T=300.15 KT=300.15\ \mathrm{K} and R=0.08206 L atm mol−1 K−1R=0.08206\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}:

M=2.62×0.08206×300.151.00≈64.5 g/mol.M=\frac{2.62\times0.08206\times300.15}{1.00} \approx64.5\ \mathrm{g/mol}.

This is close to the molar mass of sulfur dioxide, about 64.1 g/mol64.1\ \mathrm{g/mol}, but density alone does not identify a gas. Measurement uncertainty and the ideal-gas approximation also matter.

4. Comparing two states of the same gas sample

If the amount of gas stays constant,

P1V1T1=P2V2T2.\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}.

A sealed container alone is not sufficient to guarantee constant nn: a reaction or condensation could change the amount of gas present. Check the conditions in the problem.

Example 2: Compression and heating

A fixed amount of gas occupies 2.0 L2.0\ \mathrm{L} at 25∘C25^\circ\mathrm{C} and 100 kPa100\ \mathrm{kPa}. It is compressed to 1.0 L1.0\ \mathrm{L} and heated to 50∘C50^\circ\mathrm{C}. Find the final pressure, assuming ideal behavior and no loss or reaction of the gas.

Convert the temperatures to 298.15 K298.15\ \mathrm{K} and 323.15 K323.15\ \mathrm{K}:

P2=P1V1T2T1V2=100×2.0×323.15298.15×1.0≈216.8 kPa.P_2=\frac{P_1V_1T_2}{T_1V_2} =\frac{100\times2.0\times323.15}{298.15\times1.0} \approx216.8\ \mathrm{kPa}.

To two significant figures, this is 2.2×102 kPa2.2\times10^2\ \mathrm{kPa}. The pressure rises by slightly more than a factor of two because the gas is both compressed and heated.

5. Check what “standard conditions” means

Do not assume that STP always means 1 atm. IUPAC defines standard conditions for gases as 273.15 K273.15\ \mathrm{K} and 100 kPa100\ \mathrm{kPa} (1 bar); some textbooks use the older convention of 273.15 K273.15\ \mathrm{K} and 1 atm.

The ideal molar volume is approximately 22.7 L/mol22.7\ \mathrm{L/mol} under the 1 bar convention and 22.4 L/mol22.4\ \mathrm{L/mol} under the 1 atm convention. At 25∘C25^\circ\mathrm{C} and 1 bar, a commonly used ambient reference, it is about 24.8 L/mol24.8\ \mathrm{L/mol}.

Use the temperature and pressure stated in the question. A memorized molar volume is only appropriate when those conditions match.

Further reading

Content reviewed: 9 September 2026.