Various calculations, ideal gas law (st-84) Different calculations At a pressure of 75 kPa, a gas sample takes up a volume of 360 ml. What volume does the sample take up at the same temperature under a pressure of 100 kPa ? The volumes a gas sample is 462 ml at 35 ℃ and 115 kPa. What 's the volume under normal conditions ? The quantity of litres of CO (g) measured under normal conditions is required to produce 1.0 kg of rust (Fe2O3) by the following masses: Fe2O3 + CO → Fe + CO2 A Mol of a gas claims at normal conditions 22.4 litres. Calculate the side length of the Mol spindle. How many pounds of air are in a classroom? Assumptions: 180 m3, 20 ℃ 100'000 Pa, air 80% N2, 20% O2

  Exercise: st-84    

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Different calculations

1. At a pressure of 75 kPa, a gas sample takes up a volume of 360 ml. What volume does the sample take up at the same temperature under a pressure of 100 kPa ?
2. The volumes a gas sample is 462 ml at 35 ℃ and 115 kPa. What 's the volume under normal conditions ?
3. The quantity of litres of CO (g) measured under normal conditions is required to produce 1.0 kg of rust (Fe₂O₃) by the following masses:
       Fe₂O₃ + CO → Fe + CO₂
4. A Mol of a gas claims at normal conditions 22.4 litres. Calculate the side length of the Mol spindle.
5. How many pounds of air are in a classroom?
       Assumptions: 180 m³, 20 ℃ 100'000 Pa, air 80% N₂, 20% O₂

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Solution:

Note that the gas equation can be formulated as follows:
p·V = n·R·T → p·V/T = n·R
Note that after this reshaping, the term n·R is on one side. the value, no matter how large p, V and T are, is always the same and the following practical equation results:
p₁·V₁/T₁ = p₂·V₂/T₂
The values 1 and 2 shall indicate that this is situation 1 and 2.

1. Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
       p₁ = 75 kPa, V₁ = 360 ml, p₂ = 100 kPa, T₂ = T₁
       V₂ = p₁·V₁·T₁/(p₂·T₁) = p₁ ·V₁/p₂
       V₂ = 75 kPa · 360 ml / 100 kPa = 270 ml
2. Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
       p₁ = 115 kPa = 115'000 Pa, V₁ = 462 ml, T₁ = 35 ℃ = 308 K
       p₂ = 103'000 Pa, T₂ = 0 ℃ = 273 K
       V₂ = p₁·V₁·T₂/(p₂·T₁)
       V₂ = 115'000 Pa · 462 ml · 273 K / (308 K · 101'000 Pa) = 466 ml
3. Balancing the reaction equation:
       1 · Fe₂O₃ + 3 ·(g) CO → 2 · Fe + 3 · CO₂
       n(Fe₂O₃) = m/M = 1000/159.6 = 6.27 mol
       n(CO) = 3 · 6.27 = 18.8 mol
       1 mol (under standard conditions) is equivalent to 22.4 L
       18.8 mol: 421 litres
4. x³ = 22.4 L = 22.4 dm³ = 0.0224 m³
       x = ∛(0.0224m³) = 0.28 m
5. Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
       p₁ = 101'300 Pa, V₁ = 22.4 L, T₁ = 0 ℃ = 273 K
       p₂ = 100'000 Pa, T₂ = 20 ℃ = 293 K
       V₂ = p₁·V₁·T₂/(p₂·T₁)
       V₂ = 101300·22.4·293/(273·100'000) = 24.35 L
       Dichte(Luft) = (0.8·28+0.2·32) / 24.35 = 1.18 g/L
       1.18 g/L = 1.18 kg/m³
       180 m³ → 212.4 kg

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