Problem Statement
Electric current is the fundamental SI electrical dimension, with the ampere (A) as its unit. Determine units for the
following quantities as combinations of fundamental SI units.
(a) Electric power
(b) Electric charge
(c) Electric potential difference
(d) Electric resistance
(e) Electric capacitance
Solution
(a) Power is power, whether it is electrical, mechanical, or otherwise. Thus, electric power has the usual units of
power:
energy J N m kg m 2
power
time s s s3
(b) Electric current is by definition the time rate of transfer of electrical charge. Thus
charge
current
time
or charge current*time A s
(you probably recall that the Coulomb is the usual derived unit of charge, defined as 1 A s)
Solution continued on next page…
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,(c) Because power is given by the product of current and electric potential,
energy
power current*electric potential
time
energy kg m2
or electrical potential
current*time A s3
(you probably recall that this is defined as the volt)
(d) Because (by Ohm’s Law) current is electric potential divided by resistance,
electrical potential kg m2
or resistance 2 3
current A s
(this is defined as the ohm)
(e) Because electric potential is electric charge divided by electric capacitance,
charge
electrical potential
electrical capacitance
charge As A 2 s4
or electrical capacitance
electrical potential kg m 2 kg m2
3
As
Solution 1.2
Problem Statement
Liquid/vapor saturation pressure Psat is often represented as a function of temperature by the Antoine equation, which
can be written in the form:
b
log10 P sat / (torr ) a
t / C c
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consent of McGraw Hill Education.
,Here, parameters a, b, and c are substance-specific constants. Suppose this equation is to be rewritten in the equivalent
form:
B
ln P sat / kPa A
T / K C
Show how the parameters in the two equations are related.
Solution
We must convert both the units and the logarithm (between base 10 and natural logarithm). We know that t in
degrees Celsius is equal to T in Kelvins minus 273.15. Also 1 kPa is equal to 7.50 torr (we might have to look up this
conversion factor). So, we have
a b
t /C c B
P sat / torr 10 7.5 • P sat / kPa 7.5exp A
T / K C
Next, we might recognize that 10 can be written as exp(ln(10)) or exp(2.303). That is how we convert from base 10 log
to natural log in general. So,
b B
exp2.303a 7.5exp A
T / K 273.15 c T / K C
Here, I have also substituted T 273.15 for t. Taking the natural log of both sides gives
b B
2.303a ln7.5 A
T / K 273.15 c T /K C
For the two functions to be equal for all values of T, each part of the functions must be the same, so we must have
A = 2.303a ln(7.5) or A = ln(10)*a ln(7.5)
B = 2.303b or B = ln(10)*b
C=c 273.15
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, Solution 1.3
Problem Statement
Table B.2 in Appendix B provides parameters for computing the vapor pressure of many substances by the Antoine
sat
equation (see Prob. 1.2). For one of these substances, prepare two plots of P versus T over the range of temperature
sat sat
for which the parameters are valid. One plot should present P on a linear scale and the other should present P on
a log scale.
Problem 1.2
Liquid/vapor saturation pressure Psat is often represented as a function of temperature by the Antoine equation, which
can be written in the form:
b
log10 P sat / (torr ) a
t / C c
Here, parameters a, b, and c are substance-specific constants. Suppose this equation is to be rewritten in the equivalent
form:
B
ln P sat / kPa A
T / K C
Show how the parameters in the two equations are related.
Solution
The point of this problem is just for you to practice evaluating and plotting a simple function. You will do many
problems over the course of the semester (and many more over the course of your career) in which the results are best
presented in graphical form. The only thing to be careful of in plotting the Antoine equation is to pay attention to the
sat
units of T and P and to whether the constants are given for use with the base 10 logarithm or the natural logarithm.
Solution continued on next page…
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consent of McGraw Hill Education.