Question

A sample of $3.00\mathrm{g}$ of ${\mathrm{SO}}_2(g)$ originally in a 5.00-L vessel at ${21}^{\circ }\mathrm{C}$ is transferred to a $10.0$ - $\mathrm{L}$ vessel at ${26}^{\circ }\mathrm{C}$. A sample of $2.35\mathrm{g}{\mathrm{N}}_2(g)$ originally in a 2.50-L vessel at ${20}^{\circ }\mathrm{C}$ is transferred to this same $10.0$ - $\mathrm{L}$ vessel. (a) What is the partial pressure of ${\mathrm{SO}}_2(g)$ in the larger container? (b) What is the partial pressure of ${\mathrm{N}}_2(\mathrm{g})$ in this vessel? (c) What is the total pressure in the vessel?

Short answer

(a) The partial pressure of SO₂ in the larger container is 1.15 atm. (b) The partial pressure of N₂ in the larger container is 2.04 atm. (c) The total pressure in the vessel is 3.19 atm.

Step-by-step solution

Calculate the moles of SO₂ and N₂

To calculate the number of moles for each gas, we need to use the given mass and molar mass of each gas: - Molar mass of SO₂ = 32.07g/mol (S) + 2 * 16.00g/mol (O) = 64.07g/mol - Molar mass of N₂ = 2 * 14.01g/mol (N) = 28.02g/mol n(SO₂) = mass(SO₂) / molar mass(SO₂) = 3.00g / 64.07g/mol = $0.0468$ moles n(N₂) = mass(N₂) / molar mass(N₂) = 2.35g / 28.02g/mol = $0.0839$ moles

Convert temperatures to Kelvin

The ideal gas law requires temperatures to be in Kelvin, so we need to convert the given temperatures: T₁(SO₂) = 21°C + 273.15 = 294.15 K T₁(N₂) = 20°C + 273.15 = 293.15 K T₂ = 26°C + 273.15 = 299.15 K

Use the ideal gas law to find initial pressures

We'll use the ideal gas law (PV = nRT) with the initial conditions for each sample: P₁(SO₂)V₁(SO₂) = n(SO₂)R * T₁(SO₂) P₁(N₂)V₁(N₂) = n(N₂)R * T₁(N₂) The value of the ideal gas constant R = 0.0821 L⋅atm/mol⋅K Solve for P₁(SO₂) and P₁(N₂): P₁(SO₂) = (0.0468mol * 0.0821 L⋅atm/mol⋅K * 294.15 K) / 5.00 L = 2.26 atm P₁(N₂) = (0.0839mol * 0.0821 L⋅atm/mol⋅K * 293.15 K) / 2.50 L = 9.09 atm

Use the ideal gas law to find final partial pressures

Now, we need to find the final partial pressures for SO₂ and N₂ after transferring to the 10.0 L vessel: P₂(SO₂) = n(SO₂)R * T₂ / V₂ P₂(N₂) = n(N₂)R * T₂ / V₂ P₂(SO₂) = (0.0468mol * 0.0821 L⋅atm/mol⋅K * 299.15 K) / 10.0 L = 1.15 atm P₂(N₂) = (0.0839mol * 0.0821 L⋅atm/mol⋅K * 299.15 K) / 10.0 L = 2.04 atm (a) The partial pressure of SO₂ in the larger container is 1.15 atm. (b) The partial pressure of N₂ in the larger container is 2.04 atm.

Calculate the total pressure

To find the total pressure, we need to sum the partial pressures of both gases: P(total) = P₂(SO₂) + P₂(N₂) = 1.15 atm + 2.04 atm = 3.19 atm (c) The total pressure in the vessel is 3.19 atm.

Key concepts

Ideal Gas Law

The ideal gas law is a crucial concept in understanding the behavior of gases under different conditions. It is represented by the equation $PV=nRT$, where $P$ stands for pressure, $V$ is volume, $n$ represents the amount of gas in moles, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin. In practical applications, this law allows us to predict how a gas will respond to changes in temperature, volume, or amount.

To solve for the partial pressures in the exercise, the ideal gas law is rearranged to find the pressure by dividing both sides by volume, leading to $P=\frac{nRT}{V}$. This direct relationship implies that increasing the amount of gas or temperature will raise the pressure, and increasing the volume will lower it, all other factors being constant.

Molar Mass

Molar mass refers to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is an essential bridge between the microscopic world of atoms and molecules to the macroscopic world where we measure substances in grams. The calculation of molar mass is straightforward: for a compound, sum the atomic masses of all the atoms in a formula unit.

For example, sulfur dioxide ($S{O}_2$) has a molar mass calculated by adding the atomic mass of one sulfur atom (32.07 g/mol) to that of two oxygen atoms (2 × 16.00 g/mol), giving 64.07 g/mol. In our exercise, knowing the molar mass of the gases allows the conversion from grams to moles, which is necessary for using the ideal gas law.

Gas Law Constants

The gas law constants, specifically the ideal gas constant $R$, are foundational in gas law calculations as they encapsulate the physical properties of gases into a single value. For the ideal gas law, $R$ has different values depending on the units used, but a common value is 0.0821 L⋅atm/mol⋅K when dealing with pressures in atmospheres and volumes in liters.

This constant ensures that the gas law equations have consistent units and makes calculations possible. When using $R$ in calculations, it's important to ensure all other units are compatible to avoid errors. In our exercise, the value of $R=0.0821L\cdot atm/mol\cdot K$ is used to calculate the final pressures of the gases.

Kelvin Temperature Conversion

The Kelvin scale is the temperature scale used in scientific measurements because it is based on absolute zero. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. This conversion is crucial because the ideal gas law requires temperature to be in Kelvin.

Temperature conversions were needed in our example problem — turning 21°C and 20°C to Kelvin by adding 273.15, resulting in 294.15 K for $S{O}_2$ and 293.15 K for $N_2$, and 26°C to 299.15 K for the final temperature of both gases. This step is vital as using Celsius would not yield accurate results because Celsius is not an absolute scale and could lead to negative volume or pressure values when plugged into the ideal gas law.