Question

1. Room temperature water (20 Degree C) at a rate 1.5 gram/second drips into a thermal insulated container filled with liquid Nitrogen at its boiling temperature (77K). (a) How many gram of liquid Nitrogen will be vaporized every second? What is the rate of change of entropy ( Nitrogen and water) during this process? (b) What would be the Vrms of the Nitrogen molecule at 77K and 20 Degree C? Lv (Nitrogen)=5.56 kJ/mol, Lf (water) =79 cal/g, C water = 1.0 cal/g degree C, Kb = 1.32 X 10-12 J/K

Answer 1

a)

Given that m =1.5grams/s

Specific heat of water is C =1.0Cal/g^oC

Mass of nitrogen gas is N2 =62.5grams

The heat lost by the water is given by =mCDeltT =1.5*1*(293-77) =324J/s

Now heat of vaporization of nitrogen for every second is given by

= (heat used to vaporize nitrogen) / (mass of nitrogen)
= (324 J/s) / (62.5 g) =5.184J/g/s

b)

The Vrms of nitrogen molecule is given by Vrms =Sqrt(3RT/M)

Where ideal gas constant is R = 8.314 J/K/mole
Temperature is = 77K
M = molecular mas of nitrogen gas in kg = 28 X 10^-3 kg

Now Vrms at 77K is given by Vrms =Sqrt(3*8.314*77/28 X 10^-3 kg ) =261.89m/s nearly 262m/s

Now If the temperature is at 20C, that means in Kelvins 273+20=293K.
At 20C, Vrms = square root (3*8.314*293 / 28*10^-3kg) = 510.88m/s or nealry 511m/s