Calculate the temperature at which a carbon dioxide molecule would have the same root-mean-square speed as an argon atom at 60.0°C.

Question

| General Chemistry 2

Accepted Answer

v_rms= $\sqrt{\frac{3 RT}{M_{k g}}}$

v_rms(Ar at 60.0°C) =

\sqrt{\frac{3 \times 8.314 \times 333}{39.95 \times 10^{- 3}}}

= 4.56 x 10² m.s^-1

v_rms(CO₂) =

\sqrt{\frac{3 RT}{M_{kg}}}

⇒ T = v_rms(CO₂)²

\frac{M_{kg CO 2}}{3 R}

⇒ T = (4.56 x 10²)² x

\frac{44.0 \times 10^{- 3}}{3 \times 8.314}

⇒ T = 367 K = 93.8°C

Exercise 9 | Properties of Gases