The1stLawofThermodynamicsLesson2
Contents
ENTHALPY
LEARNING OUTCOMES

LESSON DEVELOPMENT
We have seen that the absorbed heat, q, in a process is not a state function. In this section we shall see how we can replace q by a state function under specified conditions. Let us first consider a process taking place at constant volume. When V = const, the pressure ¬volume work,  PdV = O. If no other work is performed, w = 0, and eq. (2.1) will take the form:
ΔU = q(2.1)
Thus q can be replaced by the change in the state function ΔU when V = const.
Let us next consider a process taking place at constant pressure, which very often is the case for chemical reactions. If no other work than pressure  volume work is performed, the work supplied when P = const, is:
where ΔV is the change in volume by the process.
Then eq. (2.1) will take the form:
ΔU = q  PΔV or q = ΔU + PΔV (2.2)
Now we shall introduce a new function, the enthalpy of a system defined by the equation:
H= U+PV(2.3)
Since U, P and V are state functions, then also H must be a state function.
For a process changing the state of a system we have:
ΔH = ΔU + Δ(PV)(2.4)
Equation (2.10) is valid for any process taking place in any system. For the specified condition P = const we have:
ΔH = ΔU + PΔV(2.5)
Comparing eqs (2.2, 2.5) we obtain:
q= ΔH(2.6)
Thus q can be replaced by the change in enthalpy, a state function, when P = const.
Internal Energy and Enthalpy for an Ideal Gas. Joule's Experiment
Internal Energy and Enthalpy for an Ideal Gas Joule's Experiment
An important experiment made by Joule in 1843 was directed at the study of the internal energy of a gas. The experimental arrangement is shown in Fig. 2.5. A closed container consisting of two bulbs connected by a valve is immersed in a calorimeter. One bulb is filled with gas, while the other bulb is evacuated. The experimental arrangement ensures that no work can be performed or supplied, w = o.
The temperature of the calorimeter is measured by means of a thermometer. After recording .the temperature the stopcock is opened, and the gas is allowed to diffuse through the valve. When equilibrium has been obtained (equal pressure in the two bulbs), the temperature is recorded again.
Joule found no temperature change for an ideal gas, ΔT = 0, which means that q = O. Since no work was performed, we have according to eq. (1.1), the first law:
ΔU = q + w = 0(2.7)
For a system where the internal energy depends on both volume and temperature,U = f(V,T), we have:
Joule's experiment shows that (∂U/∂V)_{v} = 0 for an ideal gas, hence:
This means that for an ideal gas the internal energy is a function of temperature only. From the definition of enthalpy, eg. (2.3), and the ideal gas equation (PV = nRl) we have:
ΔH = ΔU + Δ(PV) = ΔU + Δ(nRT)(2.10)
for an ideal gas, H Since the experiment showed no change in temperature, and since ΔU = 0, we have:
ΔH= 0(2.11)
for a change in pressure at constant temperature. For a system where the enthalpy depends on both pressure and temperature, H = f(P,T), we have:
For the ideal gas, H does not change with a change in P at constant T, hence: