CHEM 130B Study Guide - Midterm Guide: John Kuriyan, Junkers J 1, Joule
Chem 130A Third Mid-term Exam. April 14, 2004. Write your name here: _________________________
Page 1 of 14
UC Berkeley. Chem 130A. Spring 2004 Third mid-term Exam. April 14, 2004
Instructor: John Kuriyan ([email protected])
Enter your name & student ID number above the line, in ink.
Sign your name above the line
Exams will be returned by the Graduate Student Instructors (GSIs) during their
sections or office hours. CIRCLE THE NAME OF THE GSI FROM WHOM YOU
WILL PICK UP YOUR EXAM:
Olga Kuchment
Olayinka Oyeyemi
Romelia Salomon
Charulatha Venkataraman
This exam totals 100 points. There are 5 questions.
(For your final grade in the course,
your score in this exam will be multiplied by 2.5)
THE TIME ALLOTTED FOR THE EXAM IS FIFTY (50) MINUTES
Show all the intermediate steps in how you work out the answers to the
questions.
This exam totals 100 points. There are 4 questions.
Grading summary:
Maximum Score Actual Score
Q1 25
Q2 15
Q3 20
Q4 20
Q5 20
TOTAL 100
Chem 130A Third Mid-term Exam. April 14, 2004. Write your name here: _________________________
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Q1. (25 Points)
Consider an isolated system containing 300 black particles and 100 white
particles, as shown below. Initially, there are 100 white particles and 100 black
particles in the left hand chamber, and 200 black particles in the right hand
chamber. The two chambers are separated by a partition that is permeable only
to black particles.
(A) Calculate the total entropy of the system under the initial conditions,
assuming that each chamber is divided into 1000 grid boxes, each of
which can contain only 1 particle, and that the particles do not interact with
each other (10 points).
Hint: S = -kb lnW. Assume that kb = 1.
The multiplicity, W, of a chamber containing N grid boxes and NA particles
of type A is given by:
lnW=N(ln N−1) −NA(ln NA
−
1)
+
(N
−
NA)(ln(N
−
NA)
−
1)
[]
For left hand side, let NA = # of black particles and NB = # of white particles, N =
total #
))!((!!
!
BABA NNNNN
N
W
+−
=
)]1800(ln800)1100(ln100)1100(ln100[)11000(ln1000)
!800!100!100
!1000
ln(ln −+−+−−−==
LEFT
W
= 5907.76 – [360.52+360.52+4547.0]
= 639.03
=
RIGHT
Wln 1000(ln1000-1) – [(200(ln200-1) + 800(ln(800-1)]
= 500.4
⇒ Total Entropy = 1139.4
Chem 130A Third Mid-term Exam. April 14, 2004. Write your name here: _________________________
Page 3 of 14
Question 1(A) Continued.
(B) Given these initial conditions, will black particles move spontaneously
from the right chamber to the left chamber?
Justify your answer by calculating the change in the total entropy of the
system when one particle is moved from the right chamber to the left
chamber. (10 points)
Move one black particle from right to left.
Now =
LEFT
Wln 1000(ln1000-1) – [100(ln100-1) + 101(ln101-1)+799(ln799-1)]
= 5907.76 – [360.52+365.13+4541.0] = 641.11
=
RIGHT
Wln 1000(ln1000-1) – [199(ln199-1) + 801(ln801-1)]
= 5907.76 – [854.37+4554.4] = 499.01
Total Entropy = 1140.12
Since the total entropy of the system increases, black particles will move
spontaneously from right to left.
Document Summary
Enter your name & student id number above the line, in ink. Exams will be returned by the graduate student instructors (gsis) during their sections or office hours. Circle the name of the gsi from whom you. There are 5 questions. (for your final grade in the course, your score in this exam will be multiplied by 2. 5) The time allotted for the exam is fifty (50) minutes. Show all the intermediate steps in how you work out the answers to the questions. Consider an isolated system containing 300 black particles and 100 white particles, as shown below. Initially, there are 100 white particles and 100 black particles in the left hand chamber, and 200 black particles in the right hand chamber. For left hand side, let na = # of black particles and nb = # of white particles, n = total #
