Physical Chemistry , 1st ed.

Because the energy changes of so many processes are measured under con-
ditions of constant pressure,the change in enthalpy for a process is usually
easier to measure than the change in internal energy.As such, although the
internal energy is the more fundamental quantity, the enthalpy is the more
common.

Example 2.8

Indicate which state function is equal to heat for each process described.

a.The ignition of a sample in a bomb calorimeter, an unyielding, heavy metal

chamber in which samples are burned for heat content analysis

b.The melting of an ice cube in a cup

c.The cooling down of the inside of a refrigerator

d.A fire in a fireplace

Solution

a.From the description, one can guess that a bomb calorimeter is a constant-

volume system; therefore, the heat generated by the ignition of a sample

equals U.

b.If the cup is exposed to the atmosphere, it is subject to the (usually) con-

stant pressure of the air and so the heat of the process is equal to H.

c.A refrigerator does not change volume as it cools food, so the loss of heat

from the inside equals U.

d.A fire in a fireplace is usually exposed to the atmosphere, so the heat gen-

erated is also a measure ofH.

Example 2.9

A piston filled with 0.0400 mole of an ideal gas expands reversibly from

50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it ab-

sorbs 208 J of heat. Calculate q,w,U, and Hfor the process.

Solution

Since 208 J of heat is going into the system, the total amount of energy is

going up by 208 J, so q+208 J. In order to calculate work, we can use

equation 2.7:

w(0.0400 mol)8.314 

mo

J

lK

(310 K) ln

5

3

0

7

.

5

0

m

m

L

L



w208 J

Since Uequals q+ w,

U+208 J 208 J

U0 J

We can use equation 2.18 to calculate H, but we need to find the initial and

final pressures so we can determine (pV). Using the ideal gas law:

pi

pi20.3 atm

(0.0400 mol)(0.08205 mLoaltmK)(310 K)



0.050 L

2.5 Enthalpy 37