Question

Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is an easily liquefied gaseous fuel. Calculate the density of butane gas at 0.857 atm and \(22^{\circ} \mathrm{C}\). Give the answer in grams per liter.

Short answer

The density of butane gas is approximately 2.057 g/L.

Step-by-step solution

List the Known Variables

We know that the pressure \( P = 0.857 \) atm, the temperature \( T = 22^{\circ} \)C, and the molecular weight of butane \( \mathrm{C}_4\mathrm{H}_{10} \) is 58.12 g/mol. Convert the temperature to Kelvin by adding 273.15 to Celsius: \( T = 22 + 273.15 = 295.15 \) K.

Choose the Ideal Gas Law Formula for Density

Recall the ideal gas law: \( PV = nRT \), where \( n \) is the number of moles, \( R = 0.0821 \) L·atm/mol·K, the ideal gas constant. To find the density \( \rho \), rearrange for mass \( m = nM \) (where \( M \) is molar mass) and use \( n = \frac{m}{M} \), resulting in \( PV = \frac{m}{M}RT \). Substitute \( m = \rho V \), giving \( \rho = \frac{PM}{RT} \).

Substitute the Values into the Density Equation

Substitute the known values into the formula for density: \( \rho = \frac{(0.857 \ \text{atm})(58.12 \ \text{g/mol})}{(0.0821 \ \text{L·atm/mol·K})(295.15 \ \text{K})} \).

Calculate the Density

Calculate the expression: \( \rho = \frac{49.801}{24.2088 } \approx 2.057 \) g/L. Thus, the density of butane is approximately 2.057 g/L.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the properties of an ideal gas. It is represented by the formula \( PV = nRT \), where \( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
This equation assumes that the gas particles have no volume and do not interact with each other, which is a good approximation under many conditions. For most calculations, the ideal gas constant \( R \) is used in the form 0.0821 L·atm/mol·K.
By manipulating the Ideal Gas Law, you can solve for different properties of the gas, such as pressure, volume, temperature, or in this case, density.

Molar Mass

Molar mass is a crucial concept when working with gases, as it provides the link between the mass of a substance and the number of moles. The molar mass of a compound is the mass of one mole of that compound, usually expressed in grams per mole (g/mol).
For example, the molar mass of butane (\( \mathrm{C}_4\mathrm{H}_{10} \)) is 58.12 g/mol.
Knowing the molar mass allows you to convert between the amount of substance in moles and its mass, which is particularly useful when using the Ideal Gas Law to find properties like density.

• The molecular formula tells you the type and number of atoms in a molecule.
• To calculate molar mass, sum the atomic masses of all atoms in the molecular formula.

Gas Density Calculation

Gas density is the mass per unit volume of a gas, typically measured in grams per liter (g/L). To calculate the density of a gas using the Ideal Gas Law, we rearrange \( PV = nRT \) to the form \( \rho = \frac{PM}{RT} \).
Here, \( \rho \) represents density, \( P \) is pressure, \( M \) is molar mass, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
This rearranged formula is practical because it directly gives the density of a gas in a specific set of conditions, such as a given temperature and pressure. For example, the formula showed how the density of butane at 0.857 atm and 295.15 K is approximately 2.057 g/L.

• Ensure that all units are consistent when plugging them into the formula.
• Use the proper value of \( R \) to match the units of pressure and volume.

Temperature Conversion to Kelvin

Temperature conversion to Kelvin is essential when working with gases, as the Ideal Gas Law requires the temperature to be in Kelvin. The Kelvin scale is an absolute temperature scale where 0 Kelvin is absolute zero, the point at which molecular motion stops.
Converting Celsius to Kelvin is straightforward: simply add 273.15 to the Celsius temperature. For example, to convert \( 22^{\circ} \text{C} \) to Kelvin, you calculate \( 22 + 273.15 = 295.15 \) K.
This conversion ensures that calculations in the Ideal Gas Law remain accurate and consistent because Kelvin uniquely measures temperature in a way compatible with physical laws.

• Remember that only Kelvin, and not Celsius or Fahrenheit, is used in scientific gas calculations.
• This conversion is vital to maintain the accuracy of the gas density calculation.