20.9.1 Isothermal Processes

An isothermal process has a constant T, so we can immediately integrate Equation 20.51 to find that the entropy change in a slow isothermal process is

Δ S_{isothermal} = \frac{Q}{T}

(20.53)

where Q is the heat absorbed or released by the process. Equation 20.53 is especially important for phase changes, which occur at a constant temperature.

EXAMPLE 20.9 The entropy change of melting ice

What is the entropy change of water when 100 g of 0°C ice slowly melt to 0°C liquid water?

MODEL This phase change is a slow isothermal process.

SOLVE The heat needed for the phase change is

Q = M L_{f} = (0.100 kg) (3.33 \times 10^{5} J / kg) = 3.33 \times 10^{4} J

where the heat of fusion of ice is taken from Table 19.3. Thus the increase in entropy of the water is

Δ S = \frac{Q}{T} = \frac{3.33 \times 10^{4} J}{273 K} = 122 J / K

REVIEW Liquid water has a higher entropy than ice even at the same temperature because molecules in the liquid have more spatial mobility than do molecules in the solid and thus more ways to disperse the thermal energy.

EXAMPLE 20.10 The multiplicity increase of melting ice

By what factor does the multiplicity increase when 100 g of 0°C ice melt to 0°C liquid water?

MODEL Example 20.9 found that the entropy increase is $Δ S = 122 J / K$ . Equation 20.50 defines entropy in terms of the multiplicity. We can reverse this and use the increase in entropy to find the increase in multiplicity.

SOLVE The definition of entropy is $S = k_{B} ln Ω$ . Thus the change in entropy is

Δ S = S_{f} - S_{i} = k_{B} ln Ω_{f} - k_{B} ln Ω_{i} = k_{B} ln (\frac{Ω_{f}}{Ω_{i}})

We can use the identity $e^{ln a} = a$ to solve for $Ω_{f} / Ω_{i}$ . First we divide by $k_{B}$ , then we make both sides of the equation a power of e:

e^{ln (Ω_{f} / Ω_{i})} = \frac{Ω_{f}}{Ω_{i}} = e^{Δ S / k_{B}} = e^{(122 J/K)/(1.38 \times 10^{- 23} J / K)} = e^{8.8 \times 10^{24}}

This is hard to comprehend, but it might make a bit more sense as a power of 10. Because $log (e) = 0.434$ , using base-10 logarithms, we can write $e = 10^{0.434}$ . You will recall that $(10^{a})^{b} = 10^{a b}$ . With this, we find that melting the ice increases the water’s multiplicity by a factor of

\frac{Ω_{f}}{Ω_{i}} = e^{8.8 \times 10^{24}} = (10^{0.434})^{8.8 \times 10^{24}} = 10^{3.8 \times 10^{24}}

REVIEW It’s almost impossible to understand how staggeringly large this number is—a consequence of the extremely large number of molecules in 100 g of water. $100 = 10^{2}$ is a 1 followed by 2 zeros. $10^{24}$ , roughly the number of molecules in the 100 g of water, is a 1 followed by 24 zeros. Thus $10^{3.8 \times 10^{24}}$ is a 1 followed by $3.8 \times 10^{24}$ zeros. In print books, the zeros in a number are about 2 mm apart. Our answer, if printed, would stretch for nearly 800,000 light years, or about 8 times the diameter of our Milky Way galaxy. When we say that one macrostate is more probable than another because it has a higher multiplicity, we don’t mean 10 times more probable or even a million times more probable but something like $10^{10^{24}}$ times more probable. It is truly the case that the less probable macrostate will never occur.