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Thermal Physics PH2001. Dr Roger Bennett [email protected] Rm. 23 Xtn. 8559. Lecture 13. Entropy changes – the universe. Imagine heating a beaker of water by putting it in contact with a reservoir whose temperature remains fixed and the system remains at constant pressure. - PowerPoint PPT Presentation
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Entropy changes – the universe
• Imagine heating a beaker of water by putting it in contact with a reservoir whose temperature remains fixed and the system remains at constant pressure.
• In lecture 11 we calculated the entropy change under different process conditions.
• However as entropy is a function of state the change in entropy is defined by the end points not the process.
• What about the reservoir itself?
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Entropy changes
• The reservoir donates heat Q = Cp(Tf - Ti) at the temperature of the reservoir Tf.
• The total entropy change of the universe (system + reservoir) can therefore be calculated.
• Plugging in Ti= 273K and Tf = 373K
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Entropy changes
• The entropy of the universe increases as we expect.
• What happens if we employ an intermediate reservoir to get the water to 50°C and then the original reservoir to 100 °C?
• Plugging in Ti= 273K, T1 = 323K and Tf = 373K
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Entropy changes
• The entropy of the universe increases as we expect but by using two reservoirs the entropy change is reduced.
• Why?• What happens if we have an infinite number of
reservoirs that track the temperature of the water?
• This was our original version in lecture 11 – we used it to calculate the reversible change in entropy.
• The total change of the entropy of the universe for the reversible process is zero.
Enthalpy• Many experiments are carried out at constant
pressure –worthwhile developing a method.• Changes in volume at constant pressure lead to
work being done on or by the system.– dU = đQR - PdV
• The heat added to the system at constant pressure can be written.– đQR = dU + PdV = d(U + PV) which is true as
undertaken at constant pressure so dP = 0.• H = U + PV, is called the Enthalpy
– dH = d(U + PV) = dU + VdP + PdV = dQR + VdP– This is a general expression, the enthalpy is a
state function with units of energy.
Enthalpy• Change in volume at constant pressure – changes
of phase, boiling, melting or changing crystal structure.
• E.g. 1 mole of CaCO3 could change crystal structure at 1 bar (105Pa) with an increase in internal energy of 210 J and a density change from 2710-2930 kgm-3.
• The change in enthalpy is:-
H = 210 + 105 (34-37) 10-6 = 209.7 J.
H is very similar to U as the PV term is small at low pressure for solids. This isn’t true at high pressure or for gases.
Measuring Entropy and Latent Heats
• Consider heat being added reversibly to a system which is undergoing a phase transition.
• At the transition temperature heat flows but the temperature remains constant. The H that flows is the latent heat. +values are endothermic –ve values exothermic.
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Measuring Entropy
• To measure the entropy assume the lowest temp. we can attain is T0 and we measure Cp as a function of T from T0 up to Tf.
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Measuring Entropy
• Cp(T) can typically be measured by calorimetry except at absolute zero as we cannot get there.
• Most non-metallic solids show Cp(T) = aT3 so we can extrapolate to find S(0).
• These tend to converge on the same value of S(T=0) so we define S(T=0).
Third Law of Thermodynamics
• If the entropy of every element in its stable state at T=0 is taken to be zero, then every substance has a positive entropy which at T=0 may become zero, and does become zero for all perfectly ordered states of condensed matter.
• Such a definition makes the entropy proportional to the size (number elements) in the system – ie entropy is extensive.
• By defining S(0) we can find entropy by experiment and define new state functions such as the Helmholtz free energy F = U-TS.
• This is central to statistical mechanics that we will develop in detail.
Third Law of Thermodynamics
• If the entropy of every element in its stable state at T=0 is taken to be zero, then every substance has a positive entropy which at T=0 may become zero, and does become zero for all perfectly ordered states of condensed matter.
• Such a definition makes the entropy proportional to the size (number elements) in the system – ie entropy is extensive.
• By defining S(0) we can find entropy by experiment and define new state functions such as the Helmholtz free energy F = U-TS.
• This is central to statistical mechanics that we will develop in detail.
Properties of Entropy
• Entropy is a state function.• Entropy increases during irreversible processes.• Entropy is unchanged during reversible processes.
E.g. reversible adiabatic expansion.• Constant Entropy processes are isentropic.• Entropy is extensive.
U extensive Constraint
V extensive Constraint
n extensive Constraint
S extensive Equilibrium
P intensive Equilibrium
T intensive Equilibrium
Expansion of an ideal gas - microscopic
• Expansion of ideal gas contained in volume V.
• U, T unchanged and no work is done nor heat flows.
• Entropy increases – what is the physical basis?
Gas in V Gas in 2V
Free expansion
• No temperature change means no change in kinetic energy distribution.
• The only physical difference is that the atoms have more space in which to move.
• We may imagine that there are more ways in which the atoms may be arranged in the larger volume.
• Statistical mechanics takes this viewpoint and analyses how many different states are possible that give rise to the same macroscopic properties.
Statistical View• The constraints on the system (U, V and n) define the
macroscopic state of the system (macrostate).• We need to know how many microscopic states
(microstates or quantum states) satisfy the macrostate.• A microstate for a system is one for which everything
that can in principle be known is known.• The number of microstates that give rise to a
macrostate is called the thermodynamic probability, , of that macrostate. (alternatively the Statistical Weight W)
• The largest thermodynamic probability dominates.• E.g. think about rolling two dice, how many ways are
there of rolling 7, 2 or 12?• The essential assumption of statistical mechanics is that
each microstate is equally likely.
Statistical View
• Boltzmann’s Hypothesis:• The Entropy is a function of the statistical weight
or thermodynamic probability: S = ø(W)• If we have two systems A and B each with entropy
SA and SB respectively. Then we expect the total entropy of the two systems to be SAB = SA + SB (extensive).
• Think about the probabilities.
• WAB = WA WB
• So SAB = ø(WA) + ø(WB) = ø(WAB) = ø(WAWB)
Statistical View
• Boltzmann’s Hypothesis:
• SAB = ø(WA) + ø(WB) = ø(WAB) = ø(WAWB)
• The only functions that behave like this are logarithms.
• S = k ln(W) Boltzmann relation
• The microscopic viewpoint thus interprets the increase in entropy for an isolated system as a consequence of the natural tendency of the system to move from a less probable to a more probable state.
Expansion of an ideal gas - microscopic
• Expansion of ideal gas contained in volume V.
• U, T unchanged and no work is done nor heat flows.
• Entropy increases – what is the physical basis?
Expansion of an ideal gas - microscopic
• Split volume into elemental cells V.• Number of ways of placing one atom in
volume is V/ V.• Number of ways of placing n atoms is
– W = (V/ V)n S = nk ln(V/V)– Is this right? Depends on the size of V.
Expansion of an ideal gas - microscopic
• Is this right? Depends on the size of V.• Yep, we only measure changes in entropy.
• Sf-Si=nk(ln (Vf/V) - ln (Vi/V))= nk ln(Vf/Vi)
• Doubling volume gives S = nk ln(2) = NR ln(2)
