Question

The ratio of the RMS speeds of $CH_4$ at $T$ and $SO_2$ at 300 K is 4: 1. The average kinetic energy per mole of $CH_4$ is

• A. 3600 cal
• B. 1200 cal
• C. 900 cal
• D. 2400 cal

Answer 1

Answer: (A)

3600 cal

Answer 2

The RMS speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. The average kinetic energy per mole is $KE_{mole} = \frac{3}{2}RT$. Molar masses: $M_{CH_4} = 16$ g/mol, $M_{SO_2} = 64$ g/mol. Ratio of RMS speeds: $\frac{v_{rms, CH_4}}{v_{rms, SO_2}} = \sqrt{\frac{T_{CH_4}}{M_{CH_4}} \cdot \frac{M_{SO_2}}{T_{SO_2}}} = 4$. Substituting values ($T_{SO_2} = 300$ K): $\sqrt{\frac{T_{CH_4}}{16} \cdot \frac{64}{300}} = 4$. Solving for $T_{CH_4}$: $\frac{T_{CH_4}}{16} \cdot \frac{64}{300} = 16 \implies T_{CH_4} = 1200$ K. Average KE per mole of $CH_4$: $KE_{mole, CH_4} = \frac{3}{2}RT_{CH_4}$. Using $R \approx 2$ cal/mol·K, $KE_{mole, CH_4} = \frac{3}{2} \times 2 \times 1200 = 3600$ cal.