Question

Compare the densities, in \(\mathrm{kg} / \mathrm{m}^{3}\), of helium and air, each at \(300 \mathrm{~K}, 100 \mathrm{kPa}\). Assume ideal gas behavior.

Short answer

Density of helium: 0.1606 kg/m^3, Density of air: 1.1614 kg/m^3

Step-by-step solution

- Use the Ideal Gas Law

The ideal gas law is given by \[ PV = nRT \] where: - P is the pressure - V is the volume - n is the number of moles - R is the ideal gas constant - T is the temperature. To find density, \ \[ \rho = \frac{m}{V} \] we need the relationship between mass (m) and volume (V).

- Express density using the Ideal Gas Law

Rearrange the ideal gas law to solve for V: \[ V = \frac{nRT}{P} \] Now substitute into the density formula: \[ \rho = \frac{m}{V} = \frac{m}{\frac{nRT}{P}} \] Recall that n (number of moles) is \[ \frac{m}{M} \] where M is molar mass. Therefore: \[ V = \frac{mRT}{MP} \] So density becomes: \[ \rho = \frac{m}{V} = \frac{mP}{mRT} = \frac{MP}{RT} \]

- Calculate the density of helium

For helium, the molar mass \(M\) is \(4.0026\) g/mol. Convert it to kg/mol by dividing by 1000: \[ M = 0.0040026 \text{ kg/mol} \] Use the values: - P = 100 kPa = 100,000 Pa - T = 300 K - R = 8.314 J/(mol K) Plug into the density formula: \[ \rho_{\text{He}} = \frac{0.0040026 \times 100000}{8.314 \times 300 } = 0.1606 \text{ kg/m}^3 \]

- Calculate the density of air

For air, the average molar mass \(M\) is approximately \(28.97\) g/mol. Convert it to kg/mol by dividing by 1000: \[ M = 0.02897 \text{ kg/mol} \] Use the same values for P, T, and R: - P = 100 kPa = 100,000 Pa - T = 300 K - R = 8.314 J/(mol K) Plug into the density formula: \[ \rho_{\text{air}} = \frac{0.02897 \times 100000}{8.314 \times 300} = 1.1614 \text{ kg/m}^3 \]

Key concepts

Density Calculation

The ideal gas law helps us understand how gases behave under various conditions. It can be used to calculate the density of a gas by rearranging its formula.
Density \( \rho = \frac{m}{V} \). For the density calculation, we focus on how mass (m) and volume (V) interact.
Using the ideal gas law, \[ PV = nRT \], we can derive that \[ V = \frac{nRT}{P} \].
Now, substituting this into the density formula, we have \[ \rho = \frac{m}{V} = \frac{m}{\frac{nRT}{P}} = \frac{mP}{nRT} \].
As a result, we get \[ \rho = \frac{MP}{RT} \], where M is the molar mass of the gas.

Helium Density

Helium is a very light gas with a molar mass of 4.0026 g/mol. For density calculations, converting it to kg/mol is essential, giving 0.0040026 kg/mol.
Next, we use the known values:

• Pressure (P) = 100 kPa = 100,000 Pa
• Temperature (T) = 300 K
• Ideal gas constant (R) = 8.314 J/(mol K)

Substitute these values into our density formula:
\[ \rho_{\text{He}} = \frac{0.0040026 \times 100000}{8.314 \times 300} = 0.1606 \text{ kg/m}^3 \].
This shows that helium has a very low density.

Air Density

Air, a mixture of mostly nitrogen and oxygen, has a higher average molar mass of approximately 28.97 g/mol. Converting it to kg/mol gives 0.02897 kg/mol.
Using the same constants:

• Pressure (P) = 100 kPa = 100,000 Pa
• Temperature (T) = 300 K
• Ideal gas constant (R) = 8.314 J/(mol K)

We substitute these values into our density formula:
\[ \rho_{\text{air}} = \frac{0.02897 \times 100000}{8.314 \times 300} = 1.1614 \text{ kg/m}^3 \].
This shows that air is significantly denser than helium.

Molar Mass

The molar mass (M) of a gas is crucial for density calculations. It measures the mass of one mole of a substance. Given in grams per mole (g/mol), it's often converted to kilograms per mole (kg/mol) for calculations.
The molar mass plays a direct role in determining a gas's density, as seen in the formula \[ \rho = \frac{MP}{RT} \].
By knowing the molar mass, you can better understand why different gases have different densities. For example, helium's low molar mass makes it much less dense than air, which has a higher molar mass.