A 3.7 g mass of a gas occupies a certain vol | vedclass

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(B) Given: $W_1 = 3.7 \ g$,$T_1 = 25 + 273 = 298 \ K$,$W_2 = 0.184 \ g$,$M_2 = 2 \ g/mol$ $(H_2)$,$T_2 = 17 + 273 = 290 \ K$.Since pressure $P$ and vo

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$41.3$

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(B) Given: $W_1 = 3.7 \ g$,$T_1 = 25 + 273 = 298 \ K$,$W_2 = 0.184 \ g$,$M_2 = 2 \ g/mol$ $(H_2)$,$T_2 = 17 + 273 = 290 \ K$.Since pressure $P$ and volume $V$ are the same,from the ideal gas equation $PV = nRT$,we have $n_1 T_1 = n_2 T_2$.Substituting $n = \frac{W}{M}$,we get $\frac{W_1}{M_1} T_1 = \frac{W_2}{M_2} T_2$.$\frac{3.7}{M_1} 	imes 298 = \frac{0.184}{2} 	imes 290$.$M_1 = \frac{3.7 	imes 298 	imes 2}{0.184 	imes 290}$.$M_1 = \frac{2205.2}{53.36} \approx 41.3 \ g/mol$.

Graph	Coordinates ($y$ vs $x$)
$(A)$ Hyperbola	$1. \ y = PV, x = V$
$(B)$ Straight line passing through origin	$2. \ y = P, x = V$
$(C)$ Horizontal line parallel to $x$-axis	$3. \ y = P, x = 1/V$

$A$ $3.7 \ g$ mass of a gas occupies a certain volume at $25 \ ^\circ C$. The same volume is occupied by $0.184 \ g$ of $H_2$ gas at the same pressure and $17 \ ^\circ C$. Calculate the molar mass of the gas.

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