The density of Neon will be highest at

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(B) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and

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$0\,^{\circ}C, 2 \ atm$

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(B) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature in Kelvin.For a given gas like Neon,$M$ and $R$ are constant,so $d \propto \frac{P}{T}$.To maximize density,we need the highest pressure $(P)$ and the lowest temperature $(T)$.Comparing the options:$A$: $STP$ $(0\,^{\circ}C, 1 \ atm)$ $\rightarrow T = 273 \ K, P = 1 \ atm$ $\rightarrow \frac{P}{T} = \frac{1}{273} \approx 0.00366$$B$: $0\,^{\circ}C, 2 \ atm$ $\rightarrow T = 273 \ K, P = 2 \ atm$ $\rightarrow \frac{P}{T} = \frac{2}{273} \approx 0.00732$$C$: $273\,^{\circ}C, 1 \ atm$ $\rightarrow T = 546 \ K, P = 1 \ atm$ $\rightarrow \frac{P}{T} = \frac{1}{546} \approx 0.00183$$D$: $273\,^{\circ}C, 2 \ atm$ $\rightarrow T = 546 \ K, P = 2 \ atm$ $\rightarrow \frac{P}{T} = \frac{2}{546} \approx 0.00366$The ratio $\frac{P}{T}$ is highest for option $B$.

$P \ (bar)$	$\frac{PV}{RT}$
$1$	$1$
$2$	$1$
$3$	$1$
$4$	$1.5$
$5$	$2.0$

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