Question

please do answer question 1(a) and (b) with proper steps and calculations.

Normal air contains 78 % N2, 20.9 % O2, 0.03 % CO2 and 1.07 % H2 by weight. Determine the molar volume, molar mass and density (g cm3) of dry air at STP. Repeat the calculations for air saturated with water vapour at 25 °C and 1 atm. Assume ideal behaviour. (Vapour pressure of H20O at 25 °C- 23.756 mm Hg) (a) (10 marks) (b) The compressibility factor, Z, for a real gas is a measure of the deviation from the ideal gas behaviour. Use the following data to plot Z against P for O2 at 0°C. P/atm 100 200 400 600 800 1000 V/L mor 22.4138 0.2077 0.1024 0.0589 0.0474 0.0421 0.0389 Give three important comments on the shape of the plot. (10 marks)

Answer 1

1. (a) Let's suppose that molar mass of air is 100g and at this mass it occupies 22.4 dm³ at STP. Then N₂ will be 78g and will occupy 78% of the total volume of 22.4 dm³ (because when density is same, then ratio of masses is the ratio of volumes) . This implies that number of moles of N₂ will be 78% of the number of moles of air i.e. 1 mole (because we have supposed that molar mass of air is 100g). This is the implication from Avogadro's law which states that the number of particles of a gas (thereby its amount) is directly proportional to its volume, provided conditions of temperature and pressure remain same.

So if air is 1 mole, N₂ will be 0.78 mole (irrespective of the fact whether 1 mole of air is 100 g or not)

Mass of 0.78 mole of N₂ = 0.78 × Molar mass of N2 = 0.78 × 28 g/mol = 21.84 g/mol

Similarly, for other gases, Mass of 0.209 mole of O₂ = 0.209 × Molar mass of O2 = 0.209 × 32 g/mol = 6.688 g/mol

Mass of 0.0003 mole of CO₂ = 0.0003 × Molar mass of CO₂ = 0.0003 × 44 g/mol = 0.0132 g/mol

Mass of 0.0107 mole of H2 = 0.0107 × Molar mass of H₂ = 0.0107 × 2 g/mol = 0.0214 g/mol

Adding all the above masses, we get

Molar mass of air = 21.84 + 6.688 + 0.0132 + 0.0214 = 28.5626 g/mol ~ 29 g/mol

Its molar volume will be 22.4 dm³ or litres at STP (Avogadro's law)

Density of dry air = mass/volume = 28.5626g/22.4 dm³ = 28.5626 g /(22.4 ×1000) cm³ ~1.3 ×10^-3 g/cm³

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Now the calculations for water-vapour saturated air at 25⁰C and 1atm pressure.

Remember, the molar volume at this temperature is 24 dm³

Number of moles of water vapors can be found by applying ideal gas equation, PV = nRT

P = 23.756 mm of Hg = 23.756 × 133.32 Pa (conversion to SI units)

Now, in air saturated with water vapors at 1 atm, around 12 % is water vapors (from CRC data)

i.e. 0.12 mole of water vapors in 1 mole of air.

The percentages of N₂ and O₂ will reduce significantly. Now 78 percent N2 will be from 88 whole units and not 100 units (as in dry air).

So percentage of N2 in air saturated with water vapor = (78/100) × 88 = 68.64

Percentage of O2 in air saturated with water vapor = (20.9/100) × 88 = 18.32

Percentage of CO2 in air saturated with water vapor = (0.03/100) ×88 = 0.0264

Percentage of H2 in air saturated with water vapor = (1.07/100) × 88 = 0.9416

The rest of the calculation remains same. The answer we get is:

Molar mass of air saturated with water vapour = 19.2192 + 5.8624 +0.011616 + 0.0018832 + 2.16 = 27.255 g/mol ~27 g/mol

Density = mass/volume = 27.255/(24×1000) = 1.1 × 10^-3 g/cm³

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1 (b)Features:

200000000 180000000 160000000 140000000 120000000 100000000 80000000 60000000 40000000 20000000 7 1 2 3 5 6 PV/RT Pressure in Pa

1. It shows that O2 is not an ideal gas.

2. Z first decreases and then increases at high pressures

3. It points to the general rule that gases tend to behave ideally at low pressures

200000000 180000000 160000000 140000000 120000000 100000000 80000000 60000000 40000000 20000000 7 1 2 3 5 6 PV/RT Pressure in Pa