Problem 74 - Chemistry: The Central Science in SI Units, Global Edition

Question

A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(g)\) originally in a \(5.00-\mathrm{L}\) vessel at \(21{ }^{\circ} \mathrm{C}\) is transferred to a \(10.0-\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C}\). A sample of \(2.35 \mathrm{~g}\) of \(\mathrm{N}_{2}(g)\) originally in a \(2.50-\mathrm{L}\) vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same \(10.0-\mathrm{L}\) vessel. \((\mathbf{a})\) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

Step-by-Step Solution

Answer

(a) The partial pressure of \(SO_2\) in the 10.0-L vessel is approximately 0.115 atm. (b) The partial pressure of \(N_2\) in the 10.0-L vessel is approximately 0.405 atm. (c) The total pressure in the vessel is approximately 0.520 atm.

Step 1: Convert masses to moles

To find the moles of each gas, we divide the given mass by the molar mass: For \(SO_2\): \(\dfrac{3.00\, g}{(32.1+2\cdot16.0) g/mol} \approx 0.0465\, mol\) For \(N_2\): \(\dfrac{2.35\, g}{(2\cdot14.0) g/mol} \approx 0.0841\, mol\)

Step 2: Find initial pressure of each gas

Using the Ideal Gas Law, we can find the initial pressure for each gas: \(P = \dfrac{nRT}{V}\) For \(SO_2\): \(P_{SO_2} = \dfrac{(0.0465\, mol)(0.0821\, L\, atm/mol\, K)(294\, K)}{5.00\, L} \approx 0.229\, atm\) For \(N_2\): \(P_{N_2} = \dfrac{(0.0841\, mol)(0.0821\, L\, atm/mol\, K)(293\, K)}{2.50\, L} \approx 0.808\, atm\)

Step 3: Use the combined gas law to find the partial pressures in the new vessel

We can use the combined gas law to find the new partial pressures in the 10.0-L vessel at 26°C (299 K): \(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\) For \(SO_2\): \(\dfrac{(0.229\, atm)(5.00\, L)}{294\, K} = \dfrac{P_{SO_2} \cdot 10.0\, L}{299\, K}\) Solve for \(P_{SO_2}\): \(P_{SO_2} \approx 0.115\, atm\) For \(N_2\): \(\dfrac{(0.808\, atm)(2.50\, L)}{293\, K} = \dfrac{P_{N_2} \cdot 10.0\, L}{299\, K}\) Solve for \(P_{N_2}\): \(P_{N_2} \approx 0.405\, atm\) (a) The partial pressure of \(SO_2\) in the 10.0-L vessel is approximately 0.115 atm. (b) The partial pressure of \(N_2\) in the 10.0-L vessel is approximately 0.405 atm.

Step 4: Calculate the total pressure

To find the total pressure, add the partial pressures of the two gases: \(P_{total} = P_{SO_2} + P_{N_2} = 0.115\, atm + 0.405\, atm \approx 0.520\, atm\) (c) The total pressure in the vessel is approximately 0.520 atm.

Key Concepts

Partial Pressure

Partial pressure, in simple terms, is the pressure that a gas in a mixture would exert if it occupied the entire volume on its own. Imagine a basketball filled with different types of gases – nitrogen, oxygen, and carbon dioxide. Each gas contributes to the total pressure inside the ball, and their individual contributions are called partial pressures.

To calculate partial pressure, you need the Ideal Gas Law: \[P = \frac{nRT}{V} \] Where:

• P is the pressure,
• n is the number of moles,
• R is the gas constant (0.0821 L atm/mol K),
• T is the temperature in Kelvin,
• and V is the volume in liters.

For our classroom example, by using the number of moles of each gas, the temperature, and the volume, we can calculate how much pressure a single type of gas is exerting in a container. This is crucial for understanding the mixture's overall behavior, especially when varying the temperature and volume.

Moles Conversion

Understanding moles conversion is a fundamental part of working with gases. Moles help us connect a gas's mass with the amount of substance present. Think of moles as a bridge between the mass of a substance and its quantity in chemical reactions.

To convert mass to moles, use the formula:\[ ext{moles} = \frac{ ext{mass in grams}}{ ext{molar mass of the substance in grams per mole}} \]

• First, find the molar mass of the substance. For \ \(SO_2\), it's the sum of the atomic masses: \(32.1\, g/mol (S) + 2\times16.0\, g/mol (O) = 64.1\, g/mol\).
• Calculate the number of moles: Divide the mass of the gas in grams by its molar mass.

For example, we converted 3.00 grams of \(SO_2\) to moles by dividing by its molar mass (64.1 g/mol), finding approximately 0.0465 moles. This step is crucial for using the Ideal Gas Law, which requires the number of moles to determine pressures and volumes.

Combined Gas Law

The Combined Gas Law is a powerful tool that combines three fundamental gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law. It helps predict how a gas will behave when subjected to changes in temperature, pressure, and volume simultaneously.

The formula for the Combined Gas Law is:\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]

• \(P_1, V_1, T_1\) are the initial pressure, volume, and temperature.
• \(P_2, V_2, T_2\) are the final pressure, volume, and temperature.
• You must use the same temperature units (usually Kelvin) for reliable results.

In our exercise, we use the Combined Gas Law to determine the new partial pressures when the gases are transferred to a larger container and subjected to temperature changes. By knowing initial conditions and final volume and temperature, we can solve for the unknown new pressure, illustrating the gas's response to environmental changes. This is immensely beneficial for predicting and understanding how real-world systems operate under varying conditions.