Question

A tank of 4 m\(^3\) contains an ideal gas mixture of 60% hydrogen and 40% nitrogen by volume at 100 kPa and 300 K. Nitrogen is added to the tank such that the composition changes to 50% nitrogen by volume, with a final temperature of 300 K. The amount of nitrogen (in kmol) to be added is ........ (rounded off to three decimal places).
Use: Universal gas constant \( R_u = 8.314 \, {kJ/kmol-K} \)

Hint:

In problems involving ideal gas mixtures, you can use the ideal gas law to calculate the initial number of moles and use mole fractions to find the amount of a particular component before and after changes in composition.

Answer 1

Let's define the initial conditions: - Volume of the tank \( V = 4 \, {m}^3 \) - Pressure \( P = 100 \, {kPa} = 100 \times 10^3 \, {Pa} \) - Temperature \( T = 300 \, {K} \) - The initial mole fraction of nitrogen \( y_{{N}_2} = 0.4 \) Now, we can calculate the total number of moles in the tank initially using the ideal gas law: \[ PV = nRT \] \[ n = \frac{PV}{RT} = \frac{100 \times 10^3 \times 4}{8.314 \times 300} = 1.609 \, {kmol} \] Thus, the total number of moles initially is \( 1.609 \, {kmol} \). Since 40% of the total gas is nitrogen initially, the number of moles of nitrogen initially is: \[ n_{{N}_2} = 0.4 \times 1.609 = 0.6436 \, {kmol} \] After nitrogen is added, the composition changes to 50% nitrogen by volume, meaning the number of moles of nitrogen is now half the total moles. Thus, the new total number of moles \( n_{{new}} \) is: \[ n_{{new}} = 2 \times n_{{N}_2} = 2 \times 0.6436 = 1.2872 \, {kmol} \] The amount of nitrogen to be added is: \[ n_{{added}} = n_{{new}} - n_{{initial}} = 1.2872 - 1.609 = 0.0316 \, {kmol} \] Thus, the amount of nitrogen to be added is approximately \( 0.0316 \, {kmol} \), which lies between 0.030 and 0.034 kmol.