The volume occupied by 1.8 ; g of water vapou | vedclass

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(D) Using the ideal gas equation: $PV = nRT$ First,calculate the number of moles $(n)$ of water vapour: $n = \frac{	ext{mass}}{	ext{molar mass}} = \

Vedclass · Accepted Answer

$5.37$

Vedclass · Answer

(D) Using the ideal gas equation: $PV = nRT$ First,calculate the number of moles $(n)$ of water vapour: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{1.8 \; g}{18 \; g \; mol^{-1}} = 0.1 \; mol$ Convert the temperature $(T)$ from Celsius to Kelvin: $T = 374 + 273 = 647 \; K$ Given pressure $(P)$ is $1 \; bar$ and gas constant $(R)$ is $0.083 \; bar \; L \; K^{-1} \; mol^{-1}$. Now,calculate the volume $(V)$: $V = \frac{nRT}{P} = \frac{0.1 \; mol 	imes 0.083 \; bar \; L \; K^{-1} \; mol^{-1} 	imes 647 \; K}{1 \; bar} = 5.37 \; L$

The volume occupied by $1.8 \; g$ of water vapour at $374^{\circ} C$ and $1 \; bar$ pressure will be $....... \; L$ : [Use $R=0.083 \; bar \; L \; K^{-1} \; mol^{-1}$]

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