Page 96 - Essentials of physical chemistry

58 Essentials of Physical Chemistry

also have q ¼ w. Now we come to the only (easy) mathematics in the problem. Note that we have
to think about the gas expanding so the work on the gas is negative,
V V
ð 2 ð 2
nRT V 2 P 1
dV ¼ nRT ln ¼ nRT ln , note use of P ext ¼ P int :
w ¼ P ext dV ¼
V
int V 1 P 2
V 1 V 1
Note we have used the inverse relationship between V and P for an ideal gas, which is appropriate
because the pressures in the problem are low. Now we can put numbers in the formula.

1 atm
w ¼ (10 mol)(1:987 cal=K mol)(273:15 K) ln ¼ 12,497 cal ¼ 52, 289 J:

0:1 atm
Here we have used 1 cal ¼ 4.184 J (a calorie is worth many jewels!) due to some amazing work by
James Prescott Joule. Joule [3] performed a careful experiment measuring the increase in tempera-
ture of water caused by a rotating a paddle wheel driven by falling weights in a measurable way.
Modern recreations of the experiment produce the value of 4.184 J=cal. Note the work on the gas is
negative. Then we use the first law equation to find the heat change.

DU ¼ q þ w ¼ 0 ¼ q 12, 497 cal ¼ q 52, 289J:

Thus we have DT ¼ 0, DU ¼ 0, q ¼ 12, 497 cal ¼ 52, 289 J ¼ w. From this example, we see that
thermodynamics has some key words that have special mathematical implications and that we have
to watch the sign conventions carefully. Now let us consider another example where we change one
of the key words from ‘‘reversible’’ to ‘‘irreversible.’’

Example
Calculate q, w, and DU for the irreversible, isothermal expansion of 10 mol of ideal gas from 1 to
0.1 atm at a constant temperature of 08C.
Once again we have DT ¼ 0 and DU ¼ 0 for the same reasons as in the previous example.
However, there is a big change in the work because now we have to assume that if the final pressure
is only 0.1 atm then the complete expansion was against only that pressure of 0.1 atm. Thus, we set
up the first law as before

V ð 2 V ð 2
2 3
P ext dV ¼ q (0:1 atm)
4 5 dV ¼ q (0:1 atm)(V 2 V 1 ):
DU ¼ 0 ¼ q þ w ¼ q þ
V 1 V 1
Careful reading of the question reveals another key set of words as ‘‘ideal gas,’’ so PV ¼ nRT.

nRT nRT P 1 P 2
Thus, DU ¼ 0 ¼ q þ w ¼ q (0:1 atm) ¼ q (0:1 atm)(10RT) and
P 2 P 1 P 2 P 1
so we have a simple expression (easy mathematics!):

0:9 atm
0 ¼ q (0:1 atm)(10 mol)(1:987 cal= K mol)(273:15 K) ¼ q 4885 cal:

(0:1 atm)(1 atm)
Thus, we have q ¼ 4885 cal ¼ 20, 438 J and w ¼ q with 0 ¼ DU.
Note in particular that while the work is still negative due to the expansion of the gas (negative
work on the gas, positive work on the environment), the absolute magnitude is much less than the
reversible case and it is generally true that jw rev ¼ w max j as far as the work is calculated. The
maximum work will be obtained by a reversible process. The sign of the work is determined by
whether the system or environment is worked on. So as promised above, the mathematics of