Question

A cylinder of volume 1 m\(^3\) contains a mixture of CO\(_2\) (20% by mol) and O\(_2\) (80% by mol) at 100 kPa and 300 K. This cylinder is connected to a 1 MPa pressure line carrying N\(_2\) at 300 K. The cylinder is filled isothermally till the pressure of gas mixture inside it becomes 500 kPa, and then the filling is stopped. The amount of N\(_2\) gas that has entered the cylinder is \(\underline{\hspace{2cm}}\) (in mole, 2 decimal places). 
The universal gas constant is 8.3145 J/(mol K).

Hint:

To solve this problem, use the ideal gas law to calculate the moles of gas and consider the change in pressure when nitrogen is added isothermally.

Answer 1

Correct Answer: 159

The total initial moles of gas in the cylinder is calculated using the ideal gas law:
\[ PV = nRT \] Given that the cylinder contains a mixture of CO\(_2\) and O\(_2\), we calculate the total moles at the initial condition of 100 kPa and 300 K.
\[ n_{total} = \frac{PV}{RT} \] where \( P = 100 \, \text{kPa} = 100 \times 10^3 \, \text{Pa} \), \( V = 1 \, \text{m}^3 \), \( R = 8.3145 \, \text{J/mol K} \), and \( T = 300 \, \text{K} \).
\[ n_{total} = \frac{(100 \times 10^3) \times 1}{8.3145 \times 300} \approx 40.05 \, \text{mol}. \] The moles of each component in the mixture are:
- CO\(_2\) moles = \( 0.2 \times 40.05 \approx 8.01 \, \text{mol} \),
- O\(_2\) moles = \( 0.8 \times 40.05 \approx 32.04 \, \text{mol} \).
Now, when nitrogen (N\(_2\)) is added, the pressure in the cylinder increases from 100 kPa to 500 kPa, keeping the temperature constant. The amount of N\(_2\) that enters the cylinder can be calculated using the ideal gas law again:
\[ n_{N_2} = \frac{(P_2 - P_1) V}{RT} \] where \( P_2 = 500 \, \text{kPa} \), \( P_1 = 100 \, \text{kPa} \), and other values are the same.
\[ n_{N_2} = \frac{(500 \times 10^3 - 100 \times 10^3) \times 1}{8.3145 \times 300} \approx 159.0 \, \text{mol}. \] Thus, the amount of N\(_2\) gas that entered the cylinder is approximately 159.00 mol.