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: \(R_u = 8.314 \, kJ/kmol-K\)

Hint:

For gas mixtures, mole fraction by volume = mole fraction by moles. Always balance the ratio after adding extra gas.

Answer 1

Correct Answer: 0.03

Step 1: Apply ideal gas law.
\[ pV = n R_u T \] \[ n_{total} = \frac{pV}{R_u T} \]

Step 2: Substitute given values.
\[ p = 100 \times 10^3 \, Pa, V = 4 \, m^3, T = 300 \, K \] \[ n_{total} = \frac{100 \times 10^3 \times 4}{8.314 \times 10^3 \times 300} \] \[ n_{total} = 0.1602 \, kmol \]

Step 3: Initial moles of each component.
Hydrogen = \(0.6 \times 0.1602 = 0.0961 \, kmol\)
Nitrogen = \(0.4 \times 0.1602 = 0.0641 \, kmol\)

Step 4: Condition after nitrogen addition.
Let extra nitrogen added = \(n_x\). Final nitrogen mole fraction = 0.5: \[ \frac{0.0641 + n_x}{0.1602 + n_x} = 0.5 \]



Step 5: Solve for \(n_x\).
\[ 0.0641 + n_x = 0.5(0.1602 + n_x) \] \[ 0.0641 + n_x = 0.0801 + 0.5n_x \] \[ 0.5n_x = 0.016 \Rightarrow n_x = 0.032 \]

Step 6: Final check.
Extra nitrogen = \(0.032 \, kmol\). \[ \boxed{0.032 \, kmol} \]