Discussion Class Solutions

Week 3: 10 – 14 February

HRW 9th ed.Ch 18: Q7, Q8, P44, P47, P50

HRW 9th ed.Ch 2: Q1, Q4, P20

Chapter 18

Q7.

a) Clockwise. Work for a cyclic process is the area inside the curve of the graph and is positive if the portion of the curve with greater pressure is performed with an increase in volume i.e. in a clockwise manner.

b) Clockwise. Since this is a cyclic process then hence and hence heat transferred is positive only if work is positive which as answered in (a) happens when the cycles are traversed in a clockwise manner

Q8.

2 has the greater value for both W and Q since the work is the area encompassed by the curve in a cyclic process and this is bigger in 2 and since his is a cyclic process then hence thus the greater W means greater Q.

P44.

44. During process AB, the system is expanding, doing work on its environment, so W 0, and since Eint 0 is given then Q = W + Eint must also be positive.

a) Q > 0.

b) W > 0.

During process BC, the system is neither expanding nor contracting. Thus,

c) W = 0.

d) The sign of Eint must be the same (by the first law of thermodynamics) as that of Q, which is given as positive.Thus, Eint > 0.

During process CA, the system is contracting.The environment is doing work on the system, which implies W 0. Also, Eint 0 because Eint = 0 (for the whole cycle) and the other values of Eint (for the other processes) were positive. Therefore, Q = W + Eint must also be negative.

e) Q < 0.

f) W < 0.

g) Eint < 0.

h) The area of a triangle is (base)(height). Applying this to the figure, we find . Since process CA involves larger negative work (it occurs at higher average pressure) than the positive work done during process AB, then the net work done during the cycle must be negative. The answer is therefore Wnet = –20 J.

P47.

(a) The change in internal energy Eint is the same for path iaf and path ibf. According to the first law of thermodynamics, Eint = Q – W, where Q is the heat absorbed and W is the work done by the system. Along iaf ,

Eint = Q – W = 50 cal – 20 cal = 30 cal.

Along ibf,

W = Q – Eint = 36 cal – 30 cal = 6.0 cal.

(b) Since the curved path is traversed from f to i the change in internal energy is –30 cal and Q = Eint + W = –30 cal – 13 cal = – 43 cal.

(d) The work Wbf for the path bf is zero, so Qbf = Eint, f – Eint, b = 40 cal – 22 cal = 18 cal.

(e) For the path ibf, Q = 36 cal so Qib = Q – Qbf = 36 cal – 18 cal = 18 cal.

P50.

a) We note that process a tob is an expansion, so W > 0 for it. Thus, Wab= +5.0 J. We are told that the change in internal energy during that process is +3.0 J, so application of the first law of thermodynamics for that process immediately yields Qab= +8.0 J.

b) The net work (+1.5 J) is the same as the net heat (Qab + Qbc + Qca), and we are told that Qca = +2.5 J. Thus we readily find Qbc= (1.5 – 8.0 – 2.5) J = 9.0 J.

Chapter 2

Q1.

(a) In the negative x-direction, v is negative at t=0

(b) In the positive x-direction, v is positive at final time and will never change sign

(d) Acceleration is given by the slope of the graph. The slope is positive hence so is accereation

(e) Slope of graph is constant hence so is acceleration

Q4.

(a) Negative

(b) Velocity is given by slope of the tangent to position as a function of time. At t=1s slope is positive hence so is velocity

(d) At t=3s the slope is negative hence so is velocity

(e) Twice. Once at t=1s and again at t=3s

P20.

We use the functional notation x(t), v(t) and a(t) and find the latter two quantities by differentiating:

with SI units understood. These expressions are used in the parts that follow.

and

(a) From , we see that the only positive value of t for which the particle is (momentarily) stopped is .

(b) From 0 = – 30t, we find a(0) = 0 (that is, it vanishes at t = 0).

(d) The acceleration a(t) = – 30t is positive for t < 0.

(e) The graphs are shown below. SI units are understood.