An apparatus consists of a 4 l flask containing nitrogen gas at 27◦c and 875 kpa, joined by a valve to a 13 l flask containing argon gas at 27◦c and 54.9 kpa. the valve is opened and the gases mix. what is the partial pressure of nitrogen after mixing? answer in units of kpa.

Explanation:

Given data is as follows.

      = 4 L,         = 875 kPa

      = (27 + 273) K = 300 K,       = 13 L

      = 54.9 kPa,      

Total volume will be 4 L + 13 L = 17 L

Hence, calculate the total pressure as follows.

                        (as T is same so it will cancel out)

                    = 205.8 kPa

Now, using ideal gas equation we will calculate the moles of nitrogen as follows.

               PV = nRT

     3500 =

          n = 1.4 mol

Moles of argon will be calculated as follows.

             PV = nRT

        713.7 =

           n = 0.28 mol

Therefore, mole fraction of nitrogen will be as follows.

       Mole fraction of

                                       =

                                       =

                                       = 0.833

Relation between partial pressure and total pressure is as follows.

where,   = partial pressure of a gas

              = mole fraction of the gas

Thus, we will calculate the partial pressure of nitrogen after mixing as follows.

            205.8 kPa =

Thus, we can conclude that the partial pressure of nitrogen after mixing is 247.05 kPa.

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.