course Phy 232
Assignment 4Ch. 18 problems 54. 57, 60, 62, 67, 75, 76, 78
54)
PV=nRT
(2.00x10^3)(3000)=n(8.314)(22.0)
6=n(8.313)(22)
6=n(182.908)
N= 0.33
57)
PV=nRT v=Pi r^2h
(28.8)(0.2536)=n(8.314)(20.0) =Pi(0.0620)^2(0.21)
(28.8x0.2536)/(8.314x20.0) =1.2076(0.21)
(7.30368)/(166.28) =0.2536
N=0.0439
60)
A. What is the final Pressure?
N=PV/PT
= (1.01x10^5)(1.50x10^-3)/(8.13)(380)
=151.95/3089.4
=0.0492mol
B. How many grams of ethane remain?
Total Mass= (0.0492)(30.1)= 1.2642g
V2=1.5L(380)/(300)=1.9L
= (1.5/1.9)= 0.8L will remain
62)
1/2m(v^2)=3/2kT K= 1.3806503(24)x1-^-23 T is constant
pV=NkT
(1.00)(9.0) = N(1.3806503(24)x10^-23)(constant)
9=N(1.3806503(24)x10^-23)
N= 2.716
67)
A.
n/v= (P=an2/V2/RT)(1-bn/v)
=((9.8x105)+(0.448)/(8.314)(127))(1-(4.29x10^-5))
=(9.8x10^5/1055.878)(0.9999571)
=928.145(0.9999571)
=928.105
B.
n/v=P/RT
=(9.80x10^5/8.314(127))
=928.14
n/v=((P+an^2/V^2)/RT)(1-bn/v)
=((9.80x10^5+(0.448)(928.14)^2/(8.314)(127))(1-4.29x10^-5)(928.14)
=(1366000/1055.878)(0.60182794)
= 778.591
C.
Van Der Waals
75)
Vrms= sqrt( v^2)av = sqrt 3kT/M = sqrt 3RT/M
1.00mm/s= sqrt (3(8.13)(300)/M)
1.00/m= sqrt 3(8.13)(300)
1.00/m= sqrt 7317
1.00/m= 85.54
M=85.54
76)
I have no idea how to achieve this answer. I have all the equations but I am not sure out to get the right answer.
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Please insert a few more lines to clarify what all the given quantities mean in terms of the situation, and their units. Also be sure the goal of every problem is either obvious from your work, or is clearly stated. If you don't understand how to solve a problem, at least discuss what you do and do not understand about the problem and the given information.