1) Integrated circuits are manufactured in vacuum chambers in which the air pres

Question

1) Integrated circuits are manufactured in vacuum chambers in which the air pressure is 6.00×1011mmofHg .

Part A What is the number density? Assume T = 19.0 C.

Part B What is the mean free path of a molecule?

2) The molecules in a six-particle gas have velocities

v 1=(20i^30j^)m/s

v 2=(10i^+90j^)m/s

v 3=(50i^+20j^)m/s

v 4=30i^m/s

v 5=(40i^40j^)m/s

v 1=(50i^40j^)m/s

Part A) Calculate v avg.

Part B) Calculate vavg.

Part C) Calculate vrms.

3) At STP, what is the total translational kinetic energy of the molecules in 2.00 mol of (a) hydrogen, (b) helium, and (c) oxygen?

Part A) Kmicro(H2)=

Part B) Kmicro(He)=

Part C) Kmicro(O2)=

4) 4.30×1023 nitrogen molecules collide with a 18.0 cm2 wall each second. Assume that the molecules all travel with a speed of 360 m/s and strike the wall head on.

Part A) What is the pressure on the wall?

Explanation / Answer

1.solution

(a) 760 mm Hg = 101325 Pa
101325 Pa * ( 6*10-11 mm Hg) / (760 mm Hg)
= 7.99*10-9 Pa


pV = nRT where n is the number of moles and R = 8.314 J/(mol K)
n/V = p/RT = (7.99*10-9 Pa) / (8.314 J/(mol K) * 292.15 K)
n/V = 3.29 x 10-12 moles per m^3 (because 1 J = 1 Pa m^3)

And then use Avogadro's number,
1 mole = 6.022 x 10^23 molecules, so
(3.29 x 10-12 mol/m^3)(6.022 x 1023/mol) = 1.98 x 1012 /m^3,

(b) Mean free path = 1/(number density x collisional cross-section)
The collisional cross-section of a typical air molecule (O2 or N2) is
about 2.8 x 10-19 m^2.

Hence, mean free path = 1 / (1.98 x 1012  m^(-3) * 2.8 x 10^(-19) m^2)
=1.80 x 106 m

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