Question

the CO2 gas reaches the fire in 3.9 seconds, at what time will the CH4 reach the fire?

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Answer 1

Graham's law of effusion:

$\frac{Rate_1}{Rate_2}= \sqrt{\frac{M_2}{M_1}} \\ Rate_1=\frac{L}{t_1} \\ Rate_2 = \frac{L}{t_2} \\ M_1(CO_2) = 44 \;g/mol \\ M_2(CH_4) = 16 \;g/mol \\ \frac{\frac{L}{t_1}}{\frac{L}{t_2}}= \sqrt{\frac{16}{44}} \\ \frac{t_2}{t_1}= 0.603 \\ t_1=3.9 \;s \\ t_2=0.603t_1 \\ = 0.603 \times 3.9 = 2.35 \;s$

Answer: 2.35 s