Different calculations, Ideal Gas law (st-84) Different calculations One of them. Gas sample takes In the case of a Pressure of 75 kPa a Volumes of 360 Milligrams a. What ? Volumes takes The Sampling In the case of The equal to Temperature of which: a Pressure of 100 kPa a ? This Volumes One of them Gas sample is 462 Milligrams In the case of 35 ℃ and 115 kPa. What ? is This Volumes In the case of Normal conditions ? How a lot Litre CO (g) In the case of Normal conditions measured, will be Required, to 1.0 kg Rust (Fe2O3) The following masses to be implemented: Fe2O3 + CO → Fe + CO2 One of them Mol One of them Gases claimed In the case of Normal conditions 22.4 Litre. Calculate The Length of pages of the Mol-Wof which:. How a lot In kilograms Air are in a Classrooms? Assumptions: 180 m3, 20 ℃ 100'000 Pa, Air 80% N2, 20% O2

  Aufgabe: st-84    

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Different calculations

1. One of them. Gas sample takes In the case of a Pressure of 75 kPa a Volumes of 360 Milligrams a. What ? Volumes takes The Sampling In the case of The equal to Temperature of which: a Pressure of 100 kPa a ?
2. This Volumes One of them Gas sample is 462 Milligrams In the case of 35 ℃ and 115 kPa. What ? is This Volumes In the case of Normal conditions ?
3. How a lot Litre CO (g) In the case of Normal conditions measured, will be Required, to 1.0 kg Rust (Fe₂O₃) The following masses to be implemented:
       Fe₂O₃ + CO → Fe + CO₂
4. One of them Mol One of them Gases claimed In the case of Normal conditions 22.4 Litre. Calculate The Length of pages of the Mol-Wof which:.
5. How a lot In kilograms Air are in a Classrooms?
       Assumptions: 180 m³, 20 ℃ 100'000 Pa, Air 80% N₂, 20% O₂

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Lösung:

Please note:, that The gas equation The following masses Reworded will be can:
p·V = n·R·T → p·V/T = n·R
Please note:, that after This Conversion The Term of office n·R on One of them Page is. There you go. It 's The Commission to constant Values Acting, remaining This means that The Value, It doesn 't matter . How Large p, V and T be may, Always Right now. and the following: Practically Equation resulted:
p₁·V₁/T₁ = p₂·V₂/T₂
The Values 1 Responsibility. 2 to be on this Indications, that It 's The Commission to The The situation 1 Responsibility. 2 Acting.

1. Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
       p₁ = 75 kPa, V₁ = 360 Milligrams, p₂ = 100 kPa, T₂ = T₁
       V₂ = p₁·V₁·T₁/(p₂·T₁) = p₁ ·V₁/p₂
       V₂ = 75 kPa · 360 Milligrams / 100 kPa = 270 Milligrams
2. Es of which: p₁·V₁/T₁ = p₂·V₂/T₂ with …
       p₁ = 115 kPa = 115'000 Pa, V₁ = 462 Milligrams, T₁ = 35 ℃ = 308 K
       p₂ = 103'000 Pa, T₂ = 0 ℃ = 273 K
       V₂ = p₁·V₁·T₂/(p₂·T₁)
       V₂ = 115'000 Pa · 462 Milligrams · 273 K / (308 K · 101'000 Pa) = 466 Milligrams
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 (In the case of Standard conditions) corresponds to 22.4 L
       18.8 mol : 421 Litre
4. x³ = 22.4 L = 22.4 dm³ = 0.0224 m³
       x = ∛(0.0224m³) = 0.28 Meters
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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