Question

The gases from a wood-burning stove are found to contain \(1.8 \%\) carbon monoxide at a temperature of \(65^{\circ} \mathrm{C}\), Express the concentration in units of \(\mu \mathrm{g} \mathrm{m}^{-3}\).

Short answer

The concentration is approximately 18228.49 \( \mu \mathrm{g} \mathrm{m}^{-3} \).

Step-by-step solution

Convert Percentage to Volume Fraction

Given percentage of carbon monoxide is 1.8%. To express this in volume fraction, divide by 100: \[\text{Volume fraction} = \frac{1.8}{100} = 0.018\]

Use Ideal Gas Equation to Find Volume

The Ideal Gas equation is \( PV = nRT \), but we need concentration in \( \mu \mathrm{g} \mathrm{m}^{-3} \). First, use \( n/V = P/RT \), where:- \( P = 101325 \) Pa (assuming 1 atm as standard pressure),- \( R = 8.314 \) \( \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \),- \( T = 65^{\circ} \mathrm{C} = 273 + 65 = 338 \) K.Solve for \( n/V \):\[\frac{n}{V} = \frac{101325}{8.314 \times 338}\approx 36.24 \ mol/m^3\]

Find Molar Mass of Carbon Monoxide

Carbon monoxide (CO) has a molar mass of "C" (12 g/mol) + "O" (16 g/mol) = 28 g/mol.

Calculate Mass Concentration

Now, use the relation for the mass concentration \( \text{(in } \mu \mathrm{g} \mathrm{m}^{-3}) = \text{volume fraction} \times \text{moles of gas per volume} \times \text{molar mass} \times \text{conversion factor} \).

Calculation

Calculate the mass concentration in \( \mu \mathrm{g} \mathrm{m}^{-3} \) using:- Volume fraction: 0.018,- Moles per volume: 36.24 \( \mathrm{mol} \mathrm{m}^{-3} \),- Molar mass of CO: 28 \( \mathrm{g} \mathrm{mol}^{-1} \),- Conversion factor: 1000000 (to convert g to \( \mu g \)).The equation is:\[C = 0.018 \times 36.24 \times 28 \times 1000000 = 18228.49 \ \mu \mathrm{g} \mathrm{m}^{-3}\]

Verification

Double-check the calculations and constants for consistency and accuracy:

Key concepts

Carbon Monoxide Concentration

Carbon monoxide (CO) is a colorless and odorless gas produced from the incomplete combustion of carbon-containing fuels. To measure its concentration within a mixture of gases, such as those from a wood-burning stove, you typically begin with a percentage value which describes how much of the total gas composition it represents. In this case, the original mixture contains 1.8% carbon monoxide. To re-express this percentage as a volume fraction, divide by 100, which results in a volume fraction of 0.018. The volume fraction is crucial for further calculations because it standardizes the amount of CO for easy manipulation and comparison. Understanding gas concentrations in terms of volume fractions is important for applications in environmental science and safety monitoring, among other areas.

Molar Mass Calculation

Molar mass is an essential parameter in chemistry that helps relate the mass of a compound to the amount of substance. It is represented in grams per mole (g/mol). For carbon monoxide, the molar mass comes from its constituent elements: carbon (C) and oxygen (O). Carbon has a molar mass of 12 g/mol and oxygen 16 g/mol. Therefore, the molar mass of carbon monoxide (CO) is calculated by summing the molar masses of these elements: 12 g/mol for carbon and 16 g/mol for oxygen, resulting in a total of 28 g/mol. This number is useful when transitioning from moles to grams in calculations, allowing for a seamless connection between mass and the number of molecules or moles present in the substance.

Temperature Conversion

Temperature conversion is vital to many chemical calculations, especially when dealing with gases. The ideal gas equation requires the temperature to be in Kelvin, a unit that starts at absolute zero and is commonly used in scientific applications. The conversion from Celsius to Kelvin is simple but critical: simply add 273 to the Celsius temperature. For example, a temperature of 65°C needs to be converted to Kelvin for use in the equation. This conversion is as follows: 65°C + 273 = 338 K. Using the correct temperature unit is crucial for ensuring the calculations within equations like the ideal gas law remain valid and accurate. It standardizes the temperature across different environments and experimental conditions.

Pressure Assumptions

In many gas-related problems, standard pressure is often assumed to simplify calculations. The standard pressure value commonly used is 1 atmosphere (atm), which is equivalent to 101,325 pascals (Pa). This assumption aligns with many scientific and real-world scenarios where conditions tend to approximate standard atmospheric pressure. In exercises involving the ideal gas law, it is typically assumed unless stated otherwise, facilitating the use of the equation: \( PV = nRT \). In this context, pressure adjusts the calculation of gas volume or concentration. Without direct measurement or detail, applying this assumption helps in achieving a solution more efficiently while also aligning with established conventions in chemistry. Recognizing this standard value is essential for carrying out calculations accurately and effectively.