Problem 74 - Chemistry: The Central Science

Question

\(10.73\) A quantity of \(\mathrm{N}_{2}\) gas originally held at \(5.25 \mathrm{~atm}\) pressure in a \(1.00\) - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5\) - \(\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25\) atm and \(26^{\circ} \mathrm{C}\) in a \(5.00\)-L container is transferred to this same container. What is the total pressure in the new container?

Step-by-Step Solution

Answer

The total pressure in the new container is \(2.319\,\mathrm{atm}\).

Step 1: Convert temperatures to Kelvin

To work with the Ideal Gas Law, we need to convert the given temperatures from Celsius to Kelvin. For N2, the temperature is given as \(26^{\circ}\mathrm{C}\). To convert it to Kelvin, we add \(273.15\) \[T_{N2} = 26 + 273.15 = 299.15\,\mathrm{K}\] For O2, the temperature is also given as \(26^{\circ}\mathrm{C}\). Therefore, the temperature in Kelvin is the same as for N2: \[T_{O2} = 26 + 273.15 = 299.15\,\mathrm{K}\]

Step 2: Calculate moles of N2 and O2 using the Ideal Gas Law

Using the Ideal Gas Law, we can find moles of N2 and O2 as follows: For N2: \[n_{N2} = \frac{P_{N2}V_{N2}}{R T_{N2}} = \frac{5.25\,\mathrm{atm} \times 1.00\,\mathrm{L}}{0.0821\,\frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \times 299.15\,\mathrm{K}}\] \[n_{N2} = 0.218\,\mathrm{mol}\] For O2: \[n_{O2} = \frac{P_{O2}V_{O2}}{R T_{O2}} = \frac{5.25\,\mathrm{atm} \times 5.00\,\mathrm{L}}{0.0821\,\frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \times 299.15\,\mathrm{K}}\] \[n_{O2} = 1.089\,\mathrm{mol}\]

Step 3: Calculate the partial pressures of N2 and O2 in the new container

Now we need to find the partial pressures of N2 and O2 in the new container by using the Ideal Gas Law again. For N2, the new container has a volume of \(12.5\,\mathrm{L}\) and temperature of \(20^{\circ} \mathrm{C}\), which is \(293.15\,\mathrm{K}\). \[P_{N2} = \frac{n_{N2}RT_{new}}{V_{new}} = \frac{0.218\,\mathrm{mol} \times 0.0821\,\frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \times 293.15\,\mathrm{K}}{12.5\,\mathrm{L}}\] \[P_{N2} = 0.388\,\mathrm{atm}\] For O2, the new container volume and temperature are the same as for N2. \[P_{O2} = \frac{n_{O2}RT_{new}}{V_{new}} = \frac{1.089\,\mathrm{mol} \times 0.0821\,\frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \times 293.15\,\mathrm{K}}{12.5\,\mathrm{L}}\] \[P_{O2} = 1.931\,\mathrm{atm}\]

Step 4: Calculate the total pressure in the new container

The total pressure in the new container is the sum of the partial pressures of N2 and O2: \[P_{total} = P_{N2} + P_{O2} = 0.388\,\mathrm{atm} + 1.931\,\mathrm{atm}\] \[P_{total} = 2.319\,\mathrm{atm}\] The total pressure in the new container is \(2.319\,\mathrm{atm}\).

Key Concepts

Understanding Partial Pressure

Partial pressure refers to the pressure that a single gas in a mixture of gases would exert if it alone occupied the entire volume. It's an important concept when dealing with gas mixtures like air. In our problem, after the nitrogen and oxygen gases are mixed in the new container, each gas exerts its own pressure. These pressures can be added together to find the total pressure of the gas mixture.

• The partial pressure of a gas in a mixture is proportional to its mole fraction in the mixture.
• Dalton's Law of Partial Pressures states that the total pressure exerted by a gas mixture is equal to the sum of the partial pressures of each individual component in the gas mixture.
• In our problem, partial pressures of N₂ and O₂ are calculated separately before adding them for the total pressure.

This concept helps simplify gas problems by handling each gas as if it operates independently from others.

Temperature Conversion: Celsius to Kelvin

When working with gas laws, such as the Ideal Gas Law, it's crucial to use temperature in Kelvin rather than Celsius. This is because the Kelvin scale starts at absolute zero, which is necessary for the Ideal Gas Law to apply correctly.

• To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
• In this exercise, both N₂ and O₂ are initially at 26°C, which converts to 299.15 K.
• When transferred to the new container, the gases are at 20°C, converting to 293.15 K.

By using Kelvin, we ensure that our calculations of gas properties are accurate and compatible with the Ideal Gas Law assumptions.

Calculating Moles with Ideal Gas Law

The number of moles (n) of a gas can be determined using the Ideal Gas Law, which is \[PV = nRT\]where \(P\) is pressure, \(V\) is volume, \(R\) is the gas constant, and \(T\) is temperature in Kelvin. By rearranging the formula, we can solve for the number of moles:
\[n = \frac{PV}{RT}\]

• For \(N_2\): Given pressure and volume, we calculated 0.218 moles.
• For \(O_2\): Using the same method, we found 1.089 moles.

Knowing the number of moles is crucial for calculating partial pressures of gases in a new volume and temperature setting.

Gas Transfer Concepts Explained

Gas transfer refers to the movement of gas molecules from one container to another, often involving a change in volume and possibly temperature. This is a key part of our problem as we analyze how the initial state of \(N_2\) and \(O_2\) changes when they are moved into a larger shared container.

• When gases are transferred to a new container, properties such as pressure, volume, and temperature can change.
• The new conditions must adhere to the Ideal Gas Law to calculate new states of each gas independently.
• Once in the new container, the gases have combined partial pressures, influencing the total pressure of the gas mixture.

Understanding gas transfer helps predict how gases interact and behave in varying conditions, maintaining balance in a new environment.