the CO2 gas reaches the fire in 3.9 seconds, at what time will the CH4 reach the fire?
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Graham's law of effusion:
"\\frac{Rate_1}{Rate_2}= \\sqrt{\\frac{M_2}{M_1}} \\\\\n\nRate_1=\\frac{L}{t_1} \\\\\n\nRate_2 = \\frac{L}{t_2} \\\\\n\nM_1(CO_2) = 44 \\;g\/mol \\\\\n\nM_2(CH_4) = 16 \\;g\/mol \\\\\n\n\\frac{\\frac{L}{t_1}}{\\frac{L}{t_2}}= \\sqrt{\\frac{16}{44}} \\\\\n\n\\frac{t_2}{t_1}= 0.603 \\\\\n\nt_1=3.9 \\;s \\\\\n\nt_2=0.603t_1 \\\\\n\n= 0.603 \\times 3.9 = 2.35 \\;s"
Answer: 2.35 s
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