Nitrogen from a cylinder is bubbled through acetone at 1.1 bar and 323 K at the rate of 2 x 10^{-4} m^{3}/min. The nitrogen, saturated with acetone vapor leaves at 1.013 bar, 308 K at the rate of 3.83 x 10^{-4} m^{3}/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 ρ_{m} = n/V = P/RT = 1.1 x 10^{5}/(8314 x 323) = 0.04096 kmol/m^{3}

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 10^{5}/(8314 x 308) = 0.03956 kmol/m^{3}

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 on: 01-May-2024

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