An ideal gas with a molar mass of 40 g mol^ | vedclass

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(B) Given:Mass of gas $(w)$ = $120 \ g$Molar mass $(M)$ = $40 \ g \ mol^{-1}$Volume $(V)$ = $20 \ L$Temperature $(T)$ = $400 \ K$Gas constant $(R)$ =

Vedclass · Accepted Answer

$4.92$

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(B) Given:Mass of gas $(w)$ = $120 \ g$Molar mass $(M)$ = $40 \ g \ mol^{-1}$Volume $(V)$ = $20 \ L$Temperature $(T)$ = $400 \ K$Gas constant $(R)$ = $0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$Step $1$: Calculate the number of moles $(n)$:$n = \frac{w}{M} = \frac{120 \ g}{40 \ g \ mol^{-1}} = 3 \ mol$Step $2$: Use the ideal gas equation $PV = nRT$ to find pressure $(P)$:$P = \frac{nRT}{V}$$P = \frac{3 \ mol 	imes 0.0821 \ L \ atm \ K^{-1} \ mol^{-1} 	imes 400 \ K}{20 \ L}$$P = \frac{3 	imes 0.0821 	imes 400}{20}$$P = 3 	imes 0.0821 	imes 20$$P = 60 	imes 0.0821 = 4.926 \ atm$Rounding to the nearest provided option,the pressure is $4.92 \ atm$.

An ideal gas with a molar mass of $40 \ g \ mol^{-1}$ has a mass of $120 \ g$. It occupies a volume of $20 \ L$ at a temperature of $400 \ K$. Using $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,the pressure of the gas is ........ $atm$.

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