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- Persamaan keadaan
V = 400 L = 400×10⁻³ m³
p = 6 atm = 6×10⁵ Pa
T = 27 ºC = 300 K
Mr = 32 gr/mol
m = __ ?
soal macam ini sangat peka pada pembulatan
kalau konversinya seperti di atas, diperoleh
p V = n R T
p V = (m/Mr) R T
m = p V Mr / (R T)
m = (6×10⁵ Pa) (400×10⁻³ m³) (32 gr/mol)r / { (8,314 J/(mol K) (300 K)}
m = 3079 gr = 3,079 kg ←
p V = n R T
p V = (m/Mr) R T
m = p V Mr / (R T)
m = (6 atm) (400 L) (32 gr/mol)r / { (0,082 K atm/(mol K) (300 K)}
m = 3122 gr = 3,122 kg ←