If you have 0.132g of NO2 at a pressure of 100kPa and a temperature of 25 Degrees Celsius, what is the volume in dm^3?? Chem a level help?
回答 (2)
2016-10-14 6:33 pm
For the NO₂ gas :
Mass, m = 0.132 g
Pressure, P = 100 kPa
Absolute temperature, T = (273 + 25) K = 298 K
Volume, V = ? dm³
Gas constant, R = 8.313 J / (mol K)
Molar mass, M = (14.0 + 16.0×2) g/mol = 46 g/mol
PV = nRT and n = m/M
Then, PV = (m/M)RT
Hence, V = mRT/(PM)
Volume, V = 0.132 × 8.314 × 298 / (100 × 46) dm³ = 0.0711 dm³
Mass, m = 0.132 g
Pressure, P = 100 kPa
Absolute temperature, T = (273 + 25) K = 298 K
Volume, V = ? dm³
Gas constant, R = 8.313 J / (mol K)
Molar mass, M = (14.0 + 16.0×2) g/mol = 46 g/mol
PV = nRT and n = m/M
Then, PV = (m/M)RT
Hence, V = mRT/(PM)
Volume, V = 0.132 × 8.314 × 298 / (100 × 46) dm³ = 0.0711 dm³
2016-10-14 6:32 pm
The volume of one mole of ideal gas at 101.3 kPa and 0C is 22.414 L ( = 22.414 dm^3). The molar mass of NO2 is 14.01g + 32.00g = 46.03g, so the 0.132 grams is 0.00287 moles. Therefore the required volume is:
(0.00287)(22.414 dm^3)*(101.3/100.0)(298.15K/273.15K)
= use calculator, maybe 0.07 dm^3.
(0.00287)(22.414 dm^3)*(101.3/100.0)(298.15K/273.15K)
= use calculator, maybe 0.07 dm^3.
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