12,g of gas occupies a volume of 4times 10^{- | vedclass

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(B) Given: Mass $w = 12\,g$,Initial Volume $V_1 = 4	imes 10^{-3}\,m^3$,Initial Temperature $T_1 = 7 + 273 = 280\,K$.Density $ho = \frac{w}{V}$. Sin

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$1400$

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(B) Given: Mass $w = 12\,g$,Initial Volume $V_1 = 4	imes 10^{-3}\,m^3$,Initial Temperature $T_1 = 7 + 273 = 280\,K$.Density $ho = \frac{w}{V}$. Since pressure $P$ is constant,from the ideal gas law $PV = nRT = \frac{w}{M}RT$,we have $P = \frac{ho RT}{M}$.Since $P$ and $M$ are constant,$ho_1 T_1 = ho_2 T_2$.Initial density $ho_1 = \frac{w}{V_1} = \frac{12}{4 	imes 10^{-3}} = 3000\,g/m^3 = 3 	imes 10^{-3}\,g/cc$.Final density $ho_2 = 6 	imes 10^{-4}\,g/cc = 0.6 	imes 10^{-3}\,g/cc$.Using $ho_1 T_1 = ho_2 T_2$:$T_2 = \frac{ho_1 T_1}{ho_2} = \frac{3 	imes 10^{-3} 	imes 280}{0.6 	imes 10^{-3}} = \frac{3 	imes 280}{0.6} = 5 	imes 280 = 1400\,K$.

$12\,g$ of gas occupies a volume of $4\times 10^{-3}\,m^3$ at a temperature of $7\,^oC$. After the gas is heated at constant pressure,its density becomes $6\times 10^{-4}\,g/cc$. Find the temperature $T$ (in $K$) to which the gas was heated.

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