Question

At \(0^{\circ} \mathrm{C}\) a \(1.0-\mathrm{L}\) flask contains \(5.0 \times 10^{-2} \mathrm{~mol} \mathrm{~N}_{2}, 1.5 \times 10^{2} \mathrm{mg}\) \(\mathrm{O}_{2}\), and \(5.0 \times 10^{21}\) molecules of \(\mathrm{NH}_{3} .\) What is the partial pressure of each gas, and what is the total pressure in the flask?

Short answer

The partial pressures of N2, O2, and NH3 in the flask are 1.120 atm, 0.105 atm, and 0.186 atm respectively, and the total pressure in the flask is 1.411 atm.

Step-by-step solution

Convert the amount of each gas into moles

First, we need to convert the given amounts of each gas into moles. We are given the moles of N2, the mass of O2, and the molecules of NH3. For N2: We already have the given moles, 5.0 x 10^-2 mol. For O2: We need to convert the given mass to moles by dividing the mass by the molar mass of O2. Molar mass of O2 = 32 g/mol mass of O2 = 1.5 x 10^2 mg = 0.15 g (converting to grams) moles of O2 = \( \frac{mass}{Molar mass} = \frac{0.15}{32} = 4.6875 \times 10^{-3} mol \) For NH3: We need to convert the number of molecules of NH3 to moles using Avogadro's number (6.022 x 10^23 1/mol). Molecules of NH3 = 5 x 10^21 moles of NH3 = \( \frac{number\: of\: molecules}{Avogadro\text{'s }\: number} = \frac{5 \times 10^{21}}{6.022 \times 10^{23}} = 8.301 \times 10^{-3}mol \)

Calculate partial pressures using the Ideal Gas Law

Now that we have the amount of each gas in moles, we can use the Ideal Gas Law formula to calculate the partial pressures (P) of each gas. The Ideal Gas Law formula is given by: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant (0.0821 L atm / mol K), and T is the temperature (in Kelvin). Next, we need to convert the temperature from Celsius to Kelvin: Temperature = 0 °C + 273.15 = 273.15 K The partial pressure of each gas can be calculated by rearranging the Ideal Gas Law formula: P = \( \frac{nRT}{V} \) For N2: P_N2 = \( \frac{(5.0 \times 10^{-2} mol) (0.0821 \frac{L \cdot atm}{mol \cdot K})(273.15 K)}{1.0 L} \) = 1.120 atm For O2: P_O2 = \( \frac{(4.6875 \times 10^{-3} mol) (0.0821 \frac{L \cdot atm}{mol \cdot K})(273.15 K)}{1.0 L} = 0.105 atm For NH3: P_NH3 = \( \frac{(8.301 \times 10^{-3} mol) (0.0821 \frac{L \cdot atm}{mol \cdot K})(273.15 K)}{1.0 L} = 0.186 atm

Calculate the total pressure

In a mixture of gases, the total pressure is the sum of the partial pressures of each gas. Now that we have the partial pressures of N2, O2, and NH3, we can calculate the total pressure by adding the partial pressures: Total Pressure = P_N2 + P_O2 + P_NH3 = 1.120 atm + 0.105 atm + 0.186 atm = 1.411 atm So, the partial pressures of N2, O2, and NH3 in the flask are 1.120 atm, 0.105 atm, and 0.186 atm respectively, and the total pressure in the flask is 1.411 atm.

Key concepts

Understanding the Ideal Gas Law

The Ideal Gas Law is a fundamental principle which relates the pressure, volume, temperature, and amount of a gas. Described by the equation \( PV = nRT \), where \( P \) represents the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

For students tackling partial pressure calculations, the Ideal Gas Law serves as a crucial tool. It helps predict how a gas will behave under different conditions and can be especially useful when dealing with gas mixtures in a confined space.

In the context of classroom exercises, solving for the pressure of an individual gas in a mixture, or its partial pressure, this law is rearranged to \( P = \frac{nRT}{V} \). When applied to each gas, this yields the partial pressures which can be added together to find the total pressure within a container. Since the pressure contribution of each gas is independent, this addition aligns with Dalton's Law of Partial Pressures.

Moles to Mass Conversion in Gas Calculations

The conversion of moles to mass and vice versa is critical in chemistry, especially when working with gases. To convert the mass of a substance to moles, you divide the mass by the molar mass (the mass of one mole) of the substance. The molar mass is numerically equal to the atomic or molecular weight expressed in grams.

Understanding this conversion allows students to translate the mass of a sample, often given in milligrams (\(mg\)) or grams (\(g\)), to moles (\(mol\)) which is integral for using the Ideal Gas Law effectively. For example, oxygen (\(O_2\)) has a molar mass of 32 g/mol, so if we have 0.15 g of \(O_2\), we divide by 32 g/mol to find the moles. This calculation is a key step delineated in many exercises that deal with gas mixtures and reactions.

Navigating Avogadro's Principle

Avogadro's number, approximately \(6.022 \times 10^{23}\) entities per mole, is a constant that represents the number of atoms, ions, or molecules in one mole of a substance. This figure is immensely useful when dealing with microscopic particles at a macroscopic level.

In gas law computations, particularly when we are given the quantity of gas in terms of individual molecules, Avogadro's number provides a bridge to convert this count to moles. As such, if a problem specifies a certain number of gas molecules, you can calculate the corresponding moles by dividing the number of molecules by Avogadro's number. This is frequently the first step before you can proceed to use the Ideal Gas Law for further calculations involving gases.