A cylindrical container of volume 4.0 × 10⁻³ m³ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is :
[Take gas constant as 8.3 J mol⁻¹ K⁻¹]

Solution

Physics sir · 4 steps

Moles of an ideal-gas mixture add up: put n = 3 in PV = nRT and keep V in cubic metres.

1. Total moles. By Dalton's law each gas exerts its own pressure, and the partial pressures add. So you can treat the mixture as one gas with n = 1 + 2 = 3 mol.

   P = nRT/V = (3 × 8.3 × 400)/(4.0 × 10⁻³)

2. Put in numbers. The numerator is 3 × 8.3 × 400 = 9960 J. Dividing by 4.0 × 10⁻³ m³ gives P = 2.49 × 10⁶ Pa.
3. Check by parts. Hydrogen alone gives 8.3 × 10⁵ Pa and carbon dioxide gives 16.6 × 10⁵ Pa. Their sum is 24.9 × 10⁵ Pa, the same answer.
4. So. P = 2.49 × 10⁶ Pa = 24.9 × 10⁵ Pa, which is option (c).

The idea to keep. For a mixture of ideal gases, the kind of molecule does not matter. Only the total number of moles does. Always put volume in m³ to get pressure in pascals.

Answer from the NTA final answer key. Solution by PrepWiser.