HOMEWORK ASSIGNMENT 1
Problems Chapter 1
1.3
Plot V along the y axis and T along the x axis. Then the processes
indicated are shown on the figure shown in class.
1.6
Use the data to calculate the temperature according to
T 0 = 273.16
PT P
T
P
PT P
100
200
300
349.37 350.33 351.28
400
352.38
As PT P goes to zero the temperature approaches a finite value according to
T 0 = T + APT P
The coefficients in this linear equation can be determined by the data for the
two smallest values of PT P
349.37 = T + 100A,
350.33 = T + 200A
which gives
T = 2(349.37) − 350.33 = 348. 41,
A=
350.33 − 349.37
= 0.009 6
100
This gives an extrapolated value for the temperature of 348. 41. A least squares
linear fit to all of the data gives
T = 348.35,
A = 0.0097
The data and the two linear fits are shown on the curve. The resulting temperatures differ by 1. 722 1 × 10−2 %.
1.9
a)
The given linear relationship is
X = aT + b
Evaluating this at the ice and steam points gives
Xi = aTi + b = b,
which gives
a=
Xs = aTs + b = aTs + Xi
Xs − Xi
,
100
1
b = Xi
so the given linear relationship gives
T =
b)
X −b
X − Xi
= 100
a
Xs − Xi
The relationship is now given to be
T = a ln X + b
Repeat the same analysis of part a)
Ti = 0 = a ln Xi + b,
to get
a=
100
,
ln (Xs /Xi )
and
T = 100
1.11
a)
b)
c)
100 = a ln Xs + b
b = −100
ln Xi
ln (Xs /Xi )
ln (Xs /Xi )
ln (Xs /Xi )
T (0 C) = T (K) − 273.15 = 77.35 − 273.15 = −195.8(0 C)
T (0 F ) = 95 T (0 C) + 32 = 95 (−195.8) + 32 = −320. 44(0 F )
T (R) = T (0 F ) + 459.67 = −320. 44 + 459.67 = 139. 23
2