Question

1. The critical temperature and pressure of water are 647.1 K and 217.7 atm. For a sample of steam just below its critical point, 600.0 K and 200.0 atm, a. Calculate the number density. b. Calculate the mean free path. Assume the diameter of a water molecule is 0.15 nm. c. Compare these results to those for air at 1 atm and 298 K. d. Calculate the average speed of water molecules under these conditions, e. Calculate the collision frequency for this system. f. Calculate the collision density for this system. g. Compare your results in parts f and g with corresponding values at 1 atm and 298 K.

Answer 1

We can calculate the number density, mean free path etc. using the critical temperature and pressure. The calculation to solve the questions are given below :
(1) Calculation of number density For number density, we need to find the density Density Pressure x Molecular Mass RIGous constant) * Temperature (K) = 200 atm x 18 g mot - = 3600 Lt 0.0821 Latm mol*4,600 R 49.26 = 7308 g 4" Number density no. of atoms per unit volume - Density & Avogardo No. Molecular mass = 73.08 g L x 6023 x 10 23 atoms mot! 18 g mot = 24+ 45 x 1023 atoms Lt [1 Litse - 1000 cm] = 24-45 x 10 20 atoms an 3 (6) Calculation of Mean free path Mean free path, l= 1 Freu(diameter of molecule's Number density Fax 22 x 10 15x107 cm) x 24145 x 10 20 atom an 0.409 x 10-6 con

(C) Calculation for air (1) Number density Number density = Density & Avogardo No Molecular mass of air | Pressure x Molecular mass) x Avogaudo No. & Rx Temperature v Molecular mass - Pressure x Avogardo No. Rx Temperature = 1 X 6.023 x 1023 x atm atoms molt 0:0821 x 298 x. Latm mot kl.k = 6.023X1023 stoms 24.47 = 0.246 X1023 atoms ['-0.246 X1020 atoms com 3 (1) Mean free path Mean full path, l = 1 va ax ( diameter of air melecule) = Number density 12 x 22 * (4х 10 40 - 8 m² x 0.246x1020 atoms an3 0.057*10-4 cm

d> Average speed of water molecules 8 + Rx Temperature Average speed Scanner with CamScanner Tx Molecular mass

V 8x 8.314 I molt K+1x 600k 3:14 x 18 g mola 1706070 m2 -2 = 840mst e) Collision frequency (2) 2= Average speed mean fuce path = 840 ms 0.409 x10 6 cm = 840 x 100 cmst 0.409 x 100 cm = 2.053 X10'st Scanned with CamScanner
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