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Answer :
Final answer:
To calculate the density of neon gas at 410 mmHg and 36.9 °C, convert the given pressure and temperature to atm and K, respectively, then apply them in the modified Ideal Gas Law equation for density. The result is approximately 0.423 g/L.
Explanation:
To calculate the density of neon gas at a specific temperature and pressure, we can use the Ideal Gas Law, expressed as PV = nRT, where P stands for pressure, V for volume, n for the number of moles, R for the ideal gas constant, and T for temperature. However, to find density (ρ), we must rearrange the equation to solve for ρ, leading to ρ = (PM) / (RT), where M is the molar mass of the gas. For neon, the molar mass is 20.2 g/mol. To carry out the calculation, we'll first need to convert the pressure from mmHg to atm and the temperature from °C to Kelvin (K). Remember that 1 atm = 760 mmHg and that to convert °C to K you add 273.15 to the Celsius temperature. The pressure in atm is 410 mmHg × (1 atm / 760 mmHg) = 0.5395 atm. The temperature in Kelvin is 36.9 °C + 273.15 = 310.05 K. With these converted values, we can use the density equation to find the density of neon gas.
Density = (0.5395 atm × 20.2 g/mol) / (0.0821 L·atm/mol·K × 310.05 K) ≈ 0.423 g/L.
The density of neon gas at 410 mmHg and 36.9 °C is approximately 0.423 g/L.
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