Calculate the weight of CH_4 in a 9 L cylind | vedclass

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Question

(C) Using the ideal gas equation: $PV = \frac{w}{M}RT$Given:$P = 16 \ atm$$V = 9 \ L$$T = 27 + 273 = 300 \ K$$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(C) Using the ideal gas equation: $PV = \frac{w}{M}RT$Given:$P = 16 \ atm$$V = 9 \ L$$T = 27 + 273 = 300 \ K$$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$$M (CH_4) = 16 \ g/mol$Substituting the values:$16 	imes 9 = \frac{w}{16} 	imes 0.0821 	imes 300$$w = \frac{16 	imes 9 	imes 16}{0.0821 	imes 300} \approx 93.8 \ g$Rounding to the nearest provided option,the correct answer is $96 \ g$.

Calculate the weight of $CH_4$ in a $9 \ L$ cylinder at $16 \ atm$ and $27 \ ^\circ C$ in grams.

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