Question

A tank contains a mixture of \(52.5 \mathrm{~g}\) of oxygen gas and \(65.1 \mathrm{~g}\) of carbon dioxide gas at \(27^{\circ} \mathrm{C}\). The total pressure in the tank is 9.21 atm. Calculate the partial pressure (in atm) of each gas in the mixture.

Short answer

The partial pressure of oxygen gas (P_O2) is 4.60 atm and the partial pressure of carbon dioxide gas (P_CO2) is 4.61 atm in the tank.

Step-by-step solution

Calculate the moles of each gas

First, we need to find the moles of each gas. To do this, we will use the molar mass of oxygen gas (O2) and carbon dioxide gas (CO2), which are 32.00 g/mol and 44.01 g/mol, respectively. For oxygen gas: Moles of O2 = (Mass of O2) / (Molar mass of O2) Moles of O2 = \( \frac{52.5 \ \mathrm{g}}{32.00 \ \mathrm{g/mol}} \) For carbon dioxide gas: Moles of CO2 = (Mass of CO2) / (Molar mass of CO2) Moles of CO2 = \( \frac{65.1 \ \mathrm{g}}{44.01 \ \mathrm{g/mol}} \)

Convert the temperature to Kelvin

The temperature given is in Celsius, but we need the temperature in Kelvin for the Ideal Gas Law calculation. To convert Celsius to Kelvin, add 273.15 to the temperature in Celsius. Temperature in Kelvin = Temperature in Celsius + 273.15 Temperature in Kelvin = 27°C + 273.15

Use the Ideal Gas Law to find the partial pressure of each gas

Using the Ideal Gas Law formula, PV = nRT, where n is the moles of the gas, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. We have: P_O2 * V = n_O2 * R * T P_CO2 * V = n_CO2 * R * T We also know that the total pressure P_total is the sum of the partial pressures: P_total = P_O2 + P_CO2 We can solve the system of equations for the two unknowns by expressing P_O2: P_O2 = P_total - P_CO2 Now substitute this expression into the Ideal Gas Law equation for oxygen gas: (P_total - P_CO2) * V = n_O2 * R * T And rearrange to solve for P_CO2: P_CO2 = P_total - (n_O2 * R * T)/V Now, substitute the values for P_total, n_O2, R, and T: P_CO2 = 9.21 atm - \(\frac{ \frac{52.5 \ \mathrm{g}}{32.00 \ \mathrm{g/mol}} \times 0.0821 \ \mathrm{L \ atm/mol \ K} \times (27 + 273.15)}{V} \) Next, substitute the values for n_CO2, R, and T in the Ideal Gas Law equation for carbon dioxide gas and solve for the partial pressure of oxygen gas: P_O2 = 9.21 atm - P_CO2

Calculate the partial pressure of each gas

Once the calculations are completed in Steps 3, the resulting values for P_O2 and P_CO2 are the partial pressures of oxygen and carbon dioxide in atm, respectively. Make sure to provide the values for the partial pressures in your final answer.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry, defined by the formula \( PV = nRT \). This equation describes the behavior of ideal gases, relating the pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of a gas to the number of moles () and the ideal gas constant (\(R\)). Here, \(R\) is generally 0.0821 L atm/mol K when working with atmospheres and liters.

• Pressure (P) refers to the force exerted by the gas molecules against the walls of the container.
• Volume (V) is the space occupied by the gas.
• Temperature (T) must always be in Kelvin for these calculations.
• Moles (n) is a measure of the quantity of gas present.

The Ideal Gas Law is employed to compute different variables when the gas obeys the ideal gas conditions, which assume no intermolecular forces and that the molecules occupy no space.
In the original exercise, you're asked to find the partial pressures of gases in a mixture. Using the Ideal Gas Law, each gas's partial pressure is determined by considering its own number of moles, the shared volume, and the same temperature.

Moles Calculation

To compute the moles of gas in a substance, it's important to understand the relationship between the mass of the substance and its molar mass. The formula to find moles is given by:\[\text{Moles} = \frac{\text{Mass of Substance}}{\text{Molar Mass}}\]This simple relationship allows us to convert mass in grams to the number of moles, which is a more relevant measure for reactions.In the problem, you calculate moles separately for both oxygen (\(\text{O}_2\)) and carbon dioxide (\(\text{CO}_2\)). Given their respective molar masses (32.00 g/mol for oxygen and 44.01 g/mol for carbon dioxide), substituting the known masses allows you to find:

• The moles of \(\text{O}_2\) as \(\frac{52.5 \, \text{g}}{32.00 \, \text{g/mol}}\).
• The moles of \(\text{CO}_2\) as \(\frac{65.1 \, \text{g}}{44.01 \, \text{g/mol}}\).

Knowing the moles is fundamental for calculating other thermodynamic properties, such as pressure, implying how many molecules are present to exert pressure on the container.

Gas Mixtures

Gas mixtures are composed of two or more different gases within the same container, and they can exert a total pressure, which is the sum of the partial pressures of each constituent gas. A helpful concept here is Dalton's Law of Partial Pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the pressures that each gas would exert if it occupied the container alone.
In the exercise, Dalton’s Law helps separate the total pressure into partial pressures for oxygen and carbon dioxide, using:\[\text{Total Pressure} = P_{\text{O}_2} + P_{\text{CO}_2}\]Each gas contributes to the total pressure proportionally to its number of moles. The ideal gas behavior allows for the calculation of partial pressures based on n, R, and T.

• The partial pressure of \(\text{O}_2\) and \(\text{CO}_2\) can be individually derived, mixing equilibrium and the Ideal Gas Law, showing how gases coexist and influence each other's behavior in a mixture.

Understanding gas mixtures is essential for applications ranging from industrial processes to natural phenomena, as they represent how different gases interact in a shared space.