Calculate the density of neon gas (in g/L) at 410 mmHg and 36.9 °C.

To calculate the density of neon gas at 410 mmHg and 36.9 °C, we first convert the pressure to atmospheres and the temperature to Kelvin, then use the ideal gas law with the molar mass of neon (20.2 g/mol) to find the density, which is approximately 0.428 g/L.

Explanation:

To calculate the density of neon gas at 410 mmHg and 36.9 °C, we can use the ideal gas law PV = nRT, where P is pressure, V is volume, n is moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the given pressure into atmospheres, because the ideal gas law uses pressure in atmospheres.

P = 410 mmHg × (1 atm / 760 mmHg) = 0.5395 atm

Next, we convert the temperature to Kelvin:

T = 36.9 °C + 273.15 = 310.05 K

Then, we use the molar mass of neon, which is 20.2 g/mol, for the calculation. The ideal gas constant R is 0.0821 L·atm·mol⁻¹·K⁻¹.

We want to find the density (d), where density is mass/volume. By rearranging the ideal gas law to solve for V (V = nRT/P) and using the mass (m) as the product of n times the molar mass (M), we get the density formula d = m/V = (nM) / (nRT/P), simplifying to d = MP / (RT).

d = (20.2 g/mol × 0.5395 atm) / (0.0821 L·atm·mol⁻¹·K⁻¹ × 310.05 K)

Now, calculate the density:

d = (20.2 × 0.5395) / (0.0821 × 310.05)

d = 10.9121 / 25.481605

d = 0.428 g/L (approximated to three significant figures)

This is the density of neon gas under the given conditions.