Nitrogen from a cylinder is bubbled through acetone at 1.1 bar and 323 K at the rate of 2 x 10-4 m3/min. The nitrogen, saturated with acetone vapor leaves at 1.013 bar, 308 K at the rate of 3.83 x 10-4 m3/min. What is the vapor pressure of acetone at 308 K?

Calculations:

Using Ideal gas law, molal flow rate of nitrogen is calculated as follows:

Molal density of nitrogen ρm = n/V = P/RT = 1.1 x 105/(8314 x 323) = 0.04096 kmol/m3

Therefore, molal flow rate = 2 x 10-4 x 0.04096 = 8.1934 x 10-6 kmol/min

Similarly the molal density of leaving gases = 1.013 x 105/(8314 x 308) = 0.03956 kmol/m3

and the molal flow rate of leaving gases = 3.83 x 10-4 x 0.03956 = 1.5151 x 10-5 kmol/min

Since the molal rate of nitrogen is not going to change,

molal flow rate of acetone in the leaving gases = 1.5151 x 10-5 - 8.1934 x 10-6 = 6.9578 x 10-6 kmol/min

Mole fraction of acetone in the leaving gases = 6.9578 x 10-6 / 1.5151 x 10-5 = 0.4592

And partial pressure of acetone in the leaving gases = 0.4592 x 1.013 = 0.4652 Bar

At saturation conditions, partial pressure of vapor is equal to its vapor pressure. Therefore, vapor pressure of acetone at 308 K = 0.4652 Bar.


Last modified: Saturday, 16 March 2024, 1:40 PM