Question

Calculate the root mean square velocity, in m/s, of carbon dioxide molecules at 100.0ºC

Answer 1

T = 100^oC = 373.15 K

R = 8.314 J/mol-K

M (CO₂) = 0.04401 kg/mol

$V_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times8.314J/mol\cdot{K}\times373.15K}{0.04401kg/mol}}=459.9m/s$

Answer: 459.9 m/s