pada keadaan normal (T=0°c dan P=1atm),2gram gas oksige(O2)dengan berat molekul Mr=32memiliku vplume sebesar (R=8,31j/mol.Kdan 1 atm=10pangkat 5N/m kubik

TKG
• persamaan keadaan

m = 2 gr
Mr = 32 gr/mol
T = 0 °C = 273 K
P = 1 atm = 10⁵ Pa
V = __?

jumlah zat
n = m / Mr
n = (2 gr) / (32 gr/mol)
n = 1/16 mol

persamaan keadaan gas ideal
P V = n R T
V = n R T / P


V = (1/16 mol) • (0,082 L atm / (mol K)• (273 K) / (1 atm)
V = 1,4 L = 1,4×10⁻³ m³

V = (1/16 mol) • (8,314 J / (mol K) • (273 K) / (10⁵ Pa)
V = 1,4×10⁻³ m³