Question

Consider a mixture of two ideal gases, X and Y, with molar masses \(M_X\) = 10 kg/kmol and \(M_Y\) = 20 kg/kmol, respectively, in a container. The total pressure in the container is 100 kPa, the total volume of the container is 10 m³, and the temperature of the contents of the container is 300 K. If the mass of gas-X in the container is 2 kg, then the mass of gas-Y in the container is ............... kg. (Rounded off to one decimal place)
Assume that the universal gas constant is 8314 J kmol⁻¹K⁻¹.

Hint:

When dealing with ideal gas mixtures, always work in terms of moles. Properties like pressure and volume are additive for moles (Dalton's Law, Amagat's Law), but not directly for mass. Ensure your gas constant (\(R\) or \(R_u\)) units match the units of pressure, volume, temperature, and moles (or mass). Using the universal gas constant \(R_u\) with kilomoles (kmol) is common in engineering problems.

Answer 1

Step 1: Understanding the Concept:
This problem involves an ideal gas mixture. According to Dalton's law, the total number of moles in a mixture of gases is the sum of the number of moles of each individual gas. We can use the ideal gas law for the entire mixture to find the total number of moles, then subtract the moles of gas X to find the moles of gas Y, and finally calculate the mass of gas Y.
Step 2: Key Formula or Approach:
1. Ideal Gas Law: Relates pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), and temperature (\(T\)):
\[ PV = n R_u T \] where \(R_u\) is the universal gas constant.
2. Number of Moles: Relates mass (\(m\)) and molar mass (\(M\)):
\[ n = \frac{m}{M} \] 3. Dalton's Law (for moles): For a mixture:
\[ n_{total} = n_X + n_Y \] Step 3: Detailed Explanation or Calculation:
Given data:
Molar mass of X, \(M_X = 10\) kg/kmol
Molar mass of Y, \(M_Y = 20\) kg/kmol
Total pressure, \(P_{total} = 100\) kPa = \(100 \times 10^3\) Pa
Total volume, \(V_{total} = 10\) m³
Temperature, \(T = 300\) K
Mass of X, \(m_X = 2\) kg
Universal gas constant, \(R_u = 8314\) J kmol⁻¹K⁻¹
First, calculate the total number of moles (\(n_{total}\)) in the container using the ideal gas law:
\[ n_{total} = \frac{P_{total} V_{total}}{R_u T} = \frac{(100 \times 10^3 \text{ Pa}) \times (10 \text{ m}^3)}{8314 \frac{\text{J}}{\text{kmol} \cdot \text{K}} \times 300 \text{ K}} \] \[ n_{total} = \frac{10^6}{2494200} \approx 0.40093 \text{ kmol} \] Next, calculate the number of moles of gas X (\(n_X\)):
\[ n_X = \frac{m_X}{M_X} = \frac{2 \text{ kg}}{10 \text{ kg/kmol}} = 0.2 \text{ kmol} \] Now, find the number of moles of gas Y (\(n_Y\)) using Dalton's law:
\[ n_Y = n_{total} - n_X = 0.40093 \text{ kmol} - 0.2 \text{ kmol} = 0.20093 \text{ kmol} \] Finally, calculate the mass of gas Y (\(m_Y\)):
\[ m_Y = n_Y \times M_Y = 0.20093 \text{ kmol} \times 20 \frac{\text{kg}}{\text{kmol}} \] \[ m_Y \approx 4.0186 \text{ kg} \] Step 4: Final Answer:
Rounding off to one decimal place, the mass of gas-Y in the container is 4.0 kg.
Step 5: Why This is Correct:
The solution correctly applies the ideal gas law to the mixture to determine the total molar quantity. By calculating the moles of the known gas (X) and subtracting it from the total, we accurately find the moles of the unknown gas (Y). Converting moles of Y to mass using its molar mass gives the final answer. The result of 4.0 kg is consistent with the given answer range of 3.9 to 4.1.