Calculating Changes in Entropy (DP IB Physics: HL): Revision Note

Calculating Changes in Entropy

• At a constant temperature T, the change in entropy on a macroscopic level can be calculated using the equation

$∆S=\frac{∆Q}{T}$

• Where:

  • ΔS = change in entropy (J K⁻¹)

  • ΔQ = heat given to or removed from the system (J)

  • T = temperature of the system (K)

• When heat is given to a system (ΔQ = positive), entropy increases

$∆S=+\frac{∆Q}{T}$

$∆Q>0$

$∆S>0$

• When heat is removed from a system (ΔQ = negative), entropy decreases

$∆S=-\frac{∆Q}{T}$

$∆Q<0$

$∆S<0$

• For a reversible process that returns the system to its original state, entropy is constant

$∆Q=0$

$∆S=0$

Entropy & Microstates

• The entropy of a system, on a microscopic level, can be calculated using the equation

$S=k_B\ln \mathrm{\Omega }$

• Where:

  • S = entropy of a system of microscopic particles (J K⁻¹)

  • k_B = the Boltzmann constant

  • Ω = the number of possible microstates of the system

• Similarly, the change in entropy when the number of microstates increases from ${\mathrm{\Omega }}_1$ to ${\mathrm{\Omega }}_2$ is given by 

$∆S=k_B\ln \frac{{\mathrm{\Omega }}_2}{{\mathrm{\Omega }}_1}$

• A microstate describes one state or possible arrangement of the particles in the system 

  • A state can be defined by any microscopic or macroscopic property that is known about the system e.g. positions or velocities of molecules, energy, volume etc.

• An example that helps illustrate this is a two-compartment container which holds N distinguishable particles (i.e. each particle can be identified individually)

• Initially, all N particles are sealed in one of two compartments

The number of possible microstates describes the number of different possible arrangements of particles in a system

• When the particles are confined to one compartment, we know the location of all the particles

  • Therefore, the number of microstates (possible arrangements) in the initial volume is ${\mathrm{\Omega }}_1=1^N=1$

  • It is always equal to 1, for example, when N = 2 or N = 4:  ${\mathrm{\Omega }}_1=1^2=1^4=1$

• Once the partition is removed, the particles can spread out and occupy either one of the two compartments

  • The number of microstates (possible arrangements) in the final volume is ${\mathrm{\Omega }}_2=2^N$

  • For example, when N = 2, the particles can be arranged 2² = 4 different ways

  • Or, when N = 4, the particles can be arranged 2⁴ = 16 different ways

• The change in the entropy is therefore:

$∆S=k_B\ln \left(\frac{{\mathrm{\Omega }}_2}{{\mathrm{\Omega }}_1}\right)=k_B\ln \left(\frac{2^N}{1^N}\right)$

$∆S=k_B\ln \left(2^N\right)$

• It follows that the number of possible microstates can be equated to macroscopic properties of the gas, such as its volume increasing from V to 2V

• As the gas expands, the space it can occupy doubles, hence it gains an amount of entropy equal to:

$∆S=N{k}_B\ln \left(2V\right)-N{k}_B\ln \left(V\right)$

$∆S=N{k}_B\ln \left(\frac{2V}{V}\right)=N{k}_B\ln \left(2\right)$

• This gives the same result as above:

$∆S=k_B\ln \left(2^N\right)$