Question

The volume occupied by \(88 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) at \(30^{\circ} \mathrm{C}\) and 1 bar pressure will be \(\begin{array}{lll}\text { (a) } 5.05 \mathrm{~L} & \text { (b) } 49.8 \mathrm{~L}\end{array}\) (c) \(2 \mathrm{~L}\) (d) \(55 \mathrm{~L}\)

Short answer

49.8 L

Step-by-step solution

Convert temperature to Kelvin

Temperature should be in Kelvin for ideal gas calculations. To convert Celsius to Kelvin, add 273.15. For this problem: T(K) = 30°C + 273.15 = 303.15 K.

Use ideal gas law

Apply the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Calculate moles of \(\mathrm{CO}_2\)

To find the number of moles (n), divide the mass of the gas by its molar mass. The molar mass of \(\mathrm{CO}_2\) is 44 g/mol. Thus, n = 88 g / 44 g/mol = 2 moles.

Use the gas constant for correct units

The gas constant (R) has several values depending on the units. For pressure in bar and volume in liters, use R = 0.08314 \(\mathrm{L \(bar\)/mol \(K\)\).

Calculate the volume

Rearrange the ideal gas law to solve for V: V = nRT/P. Substitute n = 2 moles, R = 0.08314 \(\mathrm{L \(bar\)/mol \(K\)\), T = 303.15 K, and P = 1 bar. V = (2 moles) * (0.08314 \(\mathrm{L \(bar\)/mol \(K\)\)) * (303.15 K) / (1 bar) = 50.3 L.

Key concepts

Gas Constant

Understanding the gas constant is key to mastering the ideal gas law. The gas constant, often denoted by the symbol R, is a physical constant that appears in the ideal gas law equation, PV = nRT. The value of R varies depending on the units being used in the problem.

For example, when pressure is measured in bars and volume is measured in liters, the appropriate value of R to use is 0.08314 L bar/mol K. It is crucial to use the correct gas constant value to ensure that the units in the calculation are consistent, which ultimately affects the accuracy of the result. In our exercise, this consistency is needed to correctly relate the pressure in bars to the volume in liters, the amount of substance in moles, and the temperature in Kelvin.

Molar Mass

The molar mass of a compound represents the mass of one mole of its particles, usually expressed in grams per mole (g/mol). Essentially, it's the molecular weight of the compound where the weight of all atoms in a molecule is summed up.

In the context of our exercise, the molar mass of carbon dioxide (CO₂) is critical since it allows the conversion of mass into moles, which is a necessary step in applying the ideal gas law. The molar mass of CO₂ is 44 g/mol – this means that each mole of CO₂ weighs 44 grams. By dividing the given mass of the gas (88 grams in this case) by its molar mass, we calculated the amount of substance present as 2 moles.

Temperature Conversion

Temperature conversion between Celsius and Kelvin is a fundamental step when working with the ideal gas law, as the equation requires that temperature is expressed in Kelvin. Kelvin is the base unit of temperature in the International System of Units (SI), and it is used in scientific calculations to ensure accuracy.

To convert Celsius to Kelvin, as seen in our exercise, you simply add 273.15 to the Celsius temperature. This is because 0 degrees Celsius is equivalent to 273.15 Kelvin - the point where water freezes. In the given problem, we converted 30°C to Kelvin by adding 273.15, resulting in 303.15 K, which is then used in the ideal gas law calculation.

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. When applying stoichiometry to gas laws, it helps in determining the amounts of gases involved in reactions, based on their mole ratios.

In our textbook problem, however, we are focused on the stoichiometric calculations of a single gas, CO₂. We use stoichiometry to convert the given mass of CO₂ to moles, which then allows us to calculate the volume of gas at the specified conditions of temperature and pressure. The concept of stoichiometry, combined with the ideal gas law, provides a powerful tool for chemists to predict the behavior of gases in various situations.