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The Second Law of Thermodynamics
Δ S univ >0 spontaneous
Δ S univ <0 nonspontaneous (spontaneous in opposite direction)
Δ S univ = 0 reversible (system is at equilibrium)

For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, q surr is a good approximation of q rev , and the second law may be stated as the following:

Δ S univ = Δ S sys + Δ S surr = Δ S sys + q surr T

We may use this equation to predict the spontaneity of a process as illustrated in [link] .

Will ice spontaneously melt?

The entropy change for the process

H 2 O ( s ) H 2 O ( l )

is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?

Solution

We can assess the spontaneity of the process by calculating the entropy change of the universe. If Δ S univ is positive, then the process is spontaneous. At both temperatures, Δ S sys = 22.1 J/K and q surr = −6.00 kJ.

At −10.00 °C (263.15 K), the following is true:

Δ S univ = Δ S sys + Δ S surr = Δ S sys + q surr T = 22.1 J/K + −6.00 × 10 3 J 263.15 K = −0.7 J/K

S univ <0, so melting is nonspontaneous ( not spontaneous) at −10.0 °C.

At 10.00 °C (283.15 K), the following is true:

Δ S univ = Δ S sys + q surr T = 22.1 J/K + −6.00 × 10 3 J 283.15 K = +0.9 J/K

S univ >0, so melting is spontaneous at 10.00 °C.

Check your learning

Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of S univ ?

Answer:

Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.

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The third law of thermodynamics

The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal ( W = 1). According to the Boltzmann equation, the entropy of this system is zero.

S = k ln W = k ln ( 1 ) = 0

This limiting condition for a system’s entropy represents the third law of thermodynamics    : the entropy of a pure, perfect crystalline substance at 0 K is zero.

We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies are given the label S 298 ° for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The standard entropy change (Δ S °)    for any process may be computed from the standard entropies of its reactant and product species like the following:

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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