The density of a gas at 27,^oC and 1, atm is | vedclass

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(D) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is mass and $M$ is molar mass.Density $d = \frac{m}{V} = \frac{PM}{RT}$.Since $P$

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$400\, K$

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(D) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is mass and $M$ is molar mass.Density $d = \frac{m}{V} = \frac{PM}{RT}$.Since $P$ and $M$ are constant,$d \propto \frac{1}{T}$.Given $T_1 = 27 + 273 = 300\, K$ and $d_1 = d$.We need $T_2$ such that $d_2 = 0.75\, d$.Using the relation $\frac{d_2}{d_1} = \frac{T_1}{T_2}$:$\frac{0.75\, d}{d} = \frac{300}{T_2}$.$0.75 = \frac{300}{T_2} \implies T_2 = \frac{300}{0.75} = 400\, K$.Converting to Celsius: $400 - 273 = 127\,^oC$. Since $400\, K$ is provided as an option,the correct choice is $D$.

The density of a gas at $27\,^oC$ and $1\, atm$ is $d$. At constant pressure,which of the following temperatures will result in a gas density of $0.75\, d$?

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