SP48,49,50,51 Gas particles are confined to a box of volume V2. Consider a small
ID: 2251867 • Letter: S
Question
SP48,49,50,51
Gas particles are confined to a box of volume V2. Consider a smaller portion of this box of volume V1. If there are only 5 gas particles in the box, what is the probability of finding all these particles in volume V1 if V1/V2=0.8?, 0.6?, 0.4?, 0.2? Now repeat the problem with 50 gas particles in the box. Consider a box of volume V divided by an imaginary partition into two parts, V1 and V2. Start with N1 molecules of gas g1 and N2 molecules of gas g2 evenly spread throughout the box. a) What is the probability that all g1 molecules are in V1 and at the same time all g2 molecules are in V2? b) Calculate the value for the probability if V1=V/3 and V2=2V/3 for N1=30 and N2 =20. A box with a membrane down the middle has 20 N2 molecules on one side and 50 O2 molecules on the other side, a) Once the membrane is broken how many molecules of N2 and of O2 on average occupy half of the box? b) Calculate the probability of finding all the nitrogen and oxygen molecules simultaneously in their original volumes, c) Calculate the change in entropy once the membrane is broker. An inventor claims to have invented four engines, each of which operates between temperature extremes of400K and 300K. Data he gives on each engine is described below: Engine A: Qin =200J, Qout = 175J, and W = 40J Engine B: Qin =500J, Qout = 200J, and W = 400J Engine C: Qin =600J, Qout = 200J, and W = 400J Engine D: Qin =100J, Qout = 90J, and W = 10J Of the first and second laws of thermodynamics, which (if either)does each engine violate?Explanation / Answer
48) all parts are equally likely so (0.8)^5 = 0.3268
(0.6)^5 = 0.0778
(0.4)^5 = 0.0102
(0.2)^5=0.00032
now with 50 particles
(0.8)^50=0.0000143
(0.6)^50 = 8.08E-12
(0.4)^50 = 1.27E-20
(0.2)^50=1.13E-35
49)
a) (V1/V)^(N1) *(V2/V)^(N2)
b) (1/3)^30*(2/3)^20=1.46E-18
50) on average 10 N2 and 25 O2
b)(1/2)^20*(1/2)^50=8.47E-22
51)
A) breaks first law since Qout+w = 215 >Qin
B) break first law since Qout + W = 600> Qin
C) max e = 1- Tc/Th = 1-300/400 = 0.25
but e of C = 400/600 >0.25 so breaks second law
D) e of D = 10/100 = 0.1 so doesnt break second
Qout + W = 100 = Qin
so doesnt break either
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