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9476 / 2027
The Gaseous State

Lesson 2 of 3

Use pV = nRT

Choose consistent units and determine a molar mass.

A-Level 9476 (2026-2027)

  • I can use pV = nRT, including relative molecular mass.Syllabus

    Syllabus 9476, 3(c). Use pV = nRT, including relative molecular mass.

Convert the units before substituting

With R = 8.31 J mol-1 K-1, pressure is in Pa and volume in m3.

The ideal-gas equation is pV = nRT. It relates pressure p, gas volume V, amount n and absolute temperature T. With R = 8.31 J mol-1 K-1, use Pa and m3, since Pa m3 = J. A value of R with different units requires the matching pressure and volume units.

Common conversions
Given quantityConvert for SI substitution
100 kPa100000 Pa
250 cm3250 × 10-6 m3 = 2.50 × 10-4 m3
1.00 dm31.00 × 10-3 m3
25.0 degrees C298.15 K; 298 K when consistent with the precision given

Worked exampleFind a gas molar mass

A 0.444 g sample occupies 250 cm3 at 100 kPa and 298 K. Treat it as ideal. Find its molar mass.

Try it first. Then show the worked stepsHide the worked steps
  1. Convert V to 2.50 × 10-4 m3 and p to 1.00 × 105 Pa.
  2. n = pV/RT = (1.00 × 105 × 2.50 × 10-4)/(8.31 × 298) = 0.0101 mol.
  3. M = m/n = 0.444/0.0101, retaining unrounded n during calculation.
Answer

M = 44.0 g mol-1 to three significant figures. The corresponding relative molecular mass Mr is 44.0 and has no units.

Try it step by step

Work it out in steps: Volume in m3, then Amount of gas, n = pV/RT, then Molar mass, M = m/n.

  1. Problem 1. A 0.320 g sample of gas occupies 250 cm3 at 100 kPa and 300 K. Treat it as ideal and use R = 8.31 J K-1 mol-1. Find its molar mass in g mol-1 to 3 significant figures.
  2. Problem 2. A 0.580 g sample of gas occupies 240 cm3 at 100 kPa and 298 K. Treat it as ideal and use R = 8.31 J K-1 mol-1. Find its molar mass in g mol-1 to 3 significant figures.
  3. Problem 3. A 1.08 g sample of gas occupies 600 cm3 at 101 kPa and 298 K. Treat it as ideal and use R = 8.31 J K-1 mol-1. Give the volume in m3, then n and M to 3 significant figures.
Show the answer

Problem 1.

Volume in m3: V = 250 x 10-6 = 2.50 x 10-4 m3, and p = 1.00 x 105 Pa.

Amount of gas, n = pV/RT: n = (1.00 x 105 x 2.50 x 10-4) / (8.31 x 300) = 0.0100 mol.

Molar mass, M = m/n: M = 0.320 / 0.0100 = 31.9 g mol-1 (keep the unrounded n), which suggests O2.

Problem 2.

Volume in m3: V = 240 x 10-6 = 2.40 x 10-4 m3, and p = 1.00 x 105 Pa.

Amount of gas, n = pV/RT: n = (1.00 x 105 x 2.40 x 10-4) / (8.31 x 298) = 0.00969 mol.

Molar mass, M = m/n: M = 0.580 / 0.00969 = 59.8 g mol-1.

Problem 3.

Volume in m3: V = 600 x 10-6 = 0.000600 m3, and p = 101 000 Pa.

Amount of gas, n = pV/RT: n = (101 000 x 0.000600) / (8.31 x 298) = 0.0245 mol.

Molar mass, M = m/n: M = 1.08 / 0.0245 = 44.1 g mol-1, which suggests CO2 (44.0).

Combining n = m/M with the gas equation gives M = mRT/pV. Alternatively, with density ρ = m/V, M = ρRT/p. If density is in kg m-3 with SI R, the resulting molar mass is in kg mol-1; multiply by 1000 to obtain g mol-1.

A fixed gas sample also obeys p1V1/T1 = p2V2/T2, provided the amount is unchanged and the ideal approximation is suitable. Do not use a memorised room-temperature molar volume when the stated pressure or temperature differs.

Check your understanding

A sealed rigid container of gas is warmed from 20 degrees C to 40 degrees C. Does the pressure double?

  1. No. The pressure stays the same, because the rigid container keeps the gas volume fixed.
  2. No. It rises 313/293 times, about 1.07, because p follows kelvin temperature.
  3. Yes. The temperature has doubled, so the pressure doubles as well.
Show the answer

No. It rises 313/293 times, about 1.07, because p follows kelvin temperature.

At fixed n and V, p is proportional to T in kelvin. 20 degrees C is 293 K and 40 degrees C is 313 K.