Three 1.0-L flasks, maintained at 308 K, are connected to each other with stopco

Question

Three 1.0-L flasks, maintained at 308 K, are connected to each other with stopcocks. Initially the stopcocks are closed. One of the flasks contains 1.7 atm of N2, the second 3.5 g of H2O, and the third, 0.70 g of ethanol, C2H6O. The vapor pressure of H2O at 308 K is 42 mmHg and that of ethanol is 102 mmHg. The stopcocks are then opened and the contents mix freely.

I used PV=nRT to find the pressures of each and then added them together. I was told I could use P1/v1=P2/V2 by a someone... Either way I got the wrong answer in mastering chemisty. Also, I'm unsure as to what I need to do with the vapor pressures.

Thanks for your help!

Explanation / Answer

Answer – We are given the three flask with 1.0 L volume. T = 308 K

Flask 1 – N2 – pressure of N2 = 1.7 atm , mass of H2O = 3.5 g , mass of ethanol = 0.70 g

The vapor pressure of H2O at 308 K = 42 mmHg , for ethanol = 102 mmHg.

Total volume = 1.0 +1.0 +1.0 L

                       = 3.0 L

Now we need to calculate the partial pressure for each using the P1V1 =P2V2

Partial pressure of N2 –

P2 = P1V1/V2

     = 1.7 atm *1.0 L / 3.0 L

      = 0.567 atm

Partial pressure of H2O –

P1 = 42 mm Hg / 760 = 0.0553 atm

P2 = 0.0553 atm * 1.0 L / 3.0 L

     = 0.0184 atm

Partial pressure of ethanol –

P1 = 102 mm Hg / 760 = 0.134 atm

P2 = 0.134 atm * 1.0 L / 3.0 L

     = 0.0447 atm

So total pressure = 0.567 + 0.0184 + 0.0447

                           = 0.630 atm

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