The density (in g/L) of xenon (Xe) gas at 540 torr and 1150 K is:

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The density of a gas can be calculated using the ideal gas law, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the temperature. To find the density, we need to rearrange the equation to solve for $n/V$ (moles per liter), which gives us the density when multiplied by the molar mass of the gas. The ideal gas constant $R$ can be used in the form of $0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}$ or $62.36 \, \text{L torr K}^{-1} \text{mol}^{-1}$ depending on the units of pressure. Given that the pressure is in torr, we use $62.36 \, \text{L torr K}^{-1} \text{mol}^{-1}$ for $R$, and convert the pressure to atm if necessary. However, in this case, we can directly use the pressure in torr. The molar mass of xenon (Xe) is approximately $131.29 \, \text{g/mol}$.Given:- $P = 540 \, \text{torr}$- $T = 1150 \, \text{K}$- $R = 62.36 \, \text{L torr K}^{-1} \text{mol}^{-1}$- Molar mass of Xe = $131.29 \, \text{g/mol}$First, convert pressure to atm if necessary for consistency with $R$'s units, but since we're using $R$ in torr, we can skip this step.The ideal gas law rearranged for density ($\rho$) is:$$\rho = \frac{PM}{RT}$$Plugging in the values:$$\rho = \frac{540 \, \text{torr} \times 131.29 \, \text{g/mol}}{62.36 \, \text{L torr K}^{-1} \text{mol}^{-1} \times 1150 \, \text{K}}$$$$\rho \approx 0.988 \, \text{g/L}$$Therefore, the density of xenon gas under the given conditions is approximately 0.988 g/L.

The Density (In G/L) of Xenon (Xe) Gas at 540