An LPG (Liquefied Petroleum Gas) cylinder wei | vedclass

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Question

(D) Mass of gas initially $= 30.0 \ kg - 15.0 \ kg = 15.0 \ kg$. Mass of gas remaining $= 24.2 \ kg - 15.0 \ kg = 9.2 \ kg$. Mass of gas used $= 15.0

Vedclass · Accepted Answer

$2.46$

Vedclass · Answer

(D) Mass of gas initially $= 30.0 \ kg - 15.0 \ kg = 15.0 \ kg$. Mass of gas remaining $= 24.2 \ kg - 15.0 \ kg = 9.2 \ kg$. Mass of gas used $= 15.0 \ kg - 9.2 \ kg = 5.8 \ kg = 5800 \ g$. Molar mass of butane $(C_4H_{10})$ $= 4 	imes 12 + 10 	imes 1 = 58 \ g/mol$. Number of moles of gas used $(n)$ $= \frac{5800 \ g}{58 \ g/mol} = 100 \ mol$. Using the ideal gas equation $PV = nRT$ at $1 \ atm$ and $27^{\circ}C$ $(300 \ K)$: $V = \frac{nRT}{P} = \frac{100 \ mol 	imes 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 300 \ K}{1 \ atm} = 2463 \ L$. Since $1 \ m^3 = 1000 \ L$, the volume in cubic metres is $\frac{2463}{1000} \ m^3 = 2.463 \ m^3$.

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