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For compression, why does a reversible process represent the minimum work required from the surroundings (wrev = wmin.)
For reversible compression, the external pressure is kept infinitesimally higher than the gas pressure; the surroundings therefore perform the minimum force necessary to compress the gas.
True or false; all irreversible processes, and thus thinking, increase the entropy of the universe according to the second law of thermodynamics, ∆Sirrev > 0.
True
How does reversible expansion differ from reversible compression?
In reversible expansion the external pressure is kept infinitesimally lower than internal pressure and represents the maximum amount of work that can be extracted from the system; In reversible compression the external pressure is infinitesimally higher than the gas pressure and represents the minimum work required from the surroundings to achieve the compression.
Why is ∆U = q + w only true for a closed system?
Relation ∆U = q + w is the First Law of Thermodynamics for a closed system (where energy can cross the system boundary as heat or work, but mass cannot). For an open system, mass flows across boundaries, bringing additional energy transport terms into the internal energy balance.
Why does internal energy depend only on temperature for an ideal gas but not for a real gas?
For a real gas, intermolecular attractions and repulsions exist, making internal energy depend on both temperature and volume/pressure
Why is the heat capacity of a gas (Cp) temperature dependant?
Cp depends on temperature because molecular vibrational modes become progressively un-frozen and thermally accessible as temperature rises.
How does a chemical reaction affect the number of thermodynamic components?
Each independent reaction reduces the number of components by 1; to determine the number of independent components, count chemically distinguishable species and subtract independent reaction relationships and additional composition constraints: 𝐶 = 𝑆 − 𝑅 − 𝑅′.
eg. 2NH₃(g) ⇌ N₂(g)+3H₂(g); 3 species: NH₃, N₂, H₂ and 1 independent reaction; c = 3-1 = 2
What can be concluded about the sign of entropy change ∆S for a spontaneous dissolution process that causes the solution to become colder?
The solution (surroundings) lost heat, meaning the system gained it; qsys > 0 and ∆H > 0 for endothermic processes. Because dissolution proceeds spontaneously, 𝛥𝐺 < 0 → ∆G = ∆H - T∆S < 0 ; since both ∆H and T are positive quantities, entropy contribution must overcome the positive enthalpy
(Dispersing ions into solution increases the overall number of accessible configurations/micro-states)
Why can’t a temperature change alone determine the sign of ∆H or ∆G?
Temperature changes alone do not directly measure 𝛥𝐻 or establish the sign of 𝛥𝐺 because these state functions depend on the specific thermodynamic constraints and energy transfers
What is the value of ∆G when, for processes where ∆H and ∆S share the same sign, the temperature is that at which the preferred direction flips?
∆G = 0 (equilibrium); ∆G = ∆H - T∆S → T∆S = ∆H → Tflip = ∆H/∆S
What are possible explanations for the observation that no chemical change occurs after mixing three miscible liquids (A, B, and C) over the course of a week?
system is at chemical equilibrium (∆rG = 0)
non-spontaneous process: all potential chemical reactions between A, B, and C have a positive Gibbs energy change (∆rG > 0)
high activation energy barrier (∆Ga or Ea): ambient thermal energy is insufficient to overcome this barrier
absence of necessary catalyst
Why does a negative Gibbs energy change (∆G < 0) indicate that a reaction is thermodynamically favorable, but it does not guarantee that the reaction will occur rapidly or detectably?
If the chemical pathway requires breaking strong covalent bonds, the reaction will possess a high activation energy barrier (∆Ga or Ea). When ambient thermal energy is insufficient to overcome this barrier, the reaction rate will be imperceptibly slow.
An increase in Gibbs energy (∆rG > 0) corresponds to what change in total entropy (∆Stotal)?
An increase in Gibbs energy tends to lower total entropy; Holding enthalpy change (∆H) and absolute temperature (T) constant, ∆G is a linear function of ∆S with a slope equal to -T;
∆G = -T(∆S) + ∆H
As ∆S increases, the subtractive term -T∆S becomes larger in magnitude, pulling ∆G down. This represents an inverse linear (opposite) relationship, rather than a reciprocal inverse proportion.
How does increasing pressure change chemical potential (µ) at constant temperature?
Increasing pressure raises the chemical potential (μ) of every phase in direct proportion to its molar volume (Vm)
For water, how does increasing pressure change melting and boiling points (Tfus and Tb)?
Increasing pressure for water;
lowers melting point; 𝜇𝑠 increases more than 𝜇𝑙 —> Tm′(P2)<Tm(P1)
raises boiling point; 𝜇𝑔 increases more than 𝜇𝑙 —> 𝑇𝑏′ (𝑃2) > 𝑇𝑏(𝑃2); retains normal behaviour
(Spontaneous processes proceed toward states of lower chemical potential µ)
What does Chemical Potential µ measure?
µ measures the thermodynamic tendency of a substance to undergo physical change or chemical reaction
For a pure substance, µ is defined as the molar Gibbs energy: Gm = µ
Why does water spontaneously freeze?
Spontaneous processes proceed toward states of lower chemical potential; ∆µ < 0
µ (s) < µ (l)
How does the slope of the solid–liquid phase boundary on a P-T phase diagram for water differ from that of typical substances?
Water has an anomalous negative slope because solid ice occupies a larger molar volume than liquid water (Vm(liquid)<Vm(solid)), resulting in ΔfusVm<0. In contrast, typical substances expand upon melting and have a positive slope because ΔfusVm=Vm(liquid)−Vm(solid)>0, as governed by the Clapeyron equation: dTdP=TfusΔfusVmΔfusHm.
At equilibrium, CaCO₃(s) ⇌ CaO(s) + CO₂(g), how many phases are present, and why?
3 phases: CaCO₃(s), CaO(s), and CO₂(g). Each distinct solid is a separate phase; all gases mixed together count as one gas phase
What to evaluate when comparing the relative densities of phases on a P-T phase diagram?
Slope of the Solid–Liquid Coexistence Line: Governed by the Clapeyron equation, the sign of the slope (dTdP) is determined solely by ΔfusVm=Vm(liquid)−Vm(solid).
Vertical Cuts (Increasing Pressure at Constant Temperature): If you draw a vertical line straight up (increasing pressure at constant T) across a phase boundary, the system transitions to the phase occupying the smaller molar volume (the denser phase) to minimize chemical potential.
Why is there no four-phase coexistence point in a pressure–temperature phase diagram for a one-component system?
Gibb’s Phase Rule; Four phases cannot mutually coexist in equilibrium for a single component, because setting p = 4 with c = 1 yields f = -1, violating the physical constraint f ≥ 0
Why do enthalpy change and internal energy change differ for gas-phase reactions?
Gas production or consumption performs expansion work against external pressure represented by the difference term ∆ngasRT
Why are constant pressure and constant volume heat capacities nearly identical for liquids and solids?
Condensed phases undergo negligible volume changes upon heating so expansion work against surroundings is virtually zero;
Cv: heat added goes mainly into increasing the internal energy of the substance.
Cp: heat added can also do expansion work (P∆V) but ∆V is minimal
Why does standard reaction enthalpy vary with temperature according to Kirchhoff Law?
Kirchhoff's law exists because changing temperature changes the enthalpies of reactants and products according to their heat capacities, and if those heat capacities differ, their enthalpy difference (∆rH) changes.
What key assumption is required when applying the integrated Kirchhoff equation over a temperature interval?
The difference in molar heat capacities (∆Cp) between products and reactants is assumed constant over the temperature range
How does the Boltzmann equation relate thermodynamic entropy to microstates?
Entropy equals Boltzmann constant times the natural logarithm of the number of accessible spatial and energetic microstates ∆S = KBlnW
What is the conceptual difference between thermodynamic driving force ∆G and kinetic rate?
∆G determines thermodynamic feasibility and equilibrium position while activation energy (Ea) barrier determines the reaction rate
How is the reversal temperature calculated for a reaction where enthalpy and entropy changes share the same sign?
Reversal temperature equals enthalpy change divided by entropy change representing the point where delta G equals zero
∆G = 0; T∆S = ∆H → Tequ = ∆H/∆S
What does Gibbs energy change physically measure for a process at constant temperature and pressure?
The maximum theoretical amount of useful non-expansion work that can be extracted from the system
∆GP,T = wnon-exp,max
How does ATP hydrolysis drive endergonic metabolic pathways in living cells?
Enzymatic coupling links favorable ATP hydrolysis to an unfavorable reaction yielding a net negative coupled Gibbs energy change
∆rGcoupled = ∆rGx(unfavourable) + ∆rGhydrolysis
What does the partial derivative of molar Gibbs energy with respect to pressure at constant temperature represent?
It equals molar volume indicating molar Gibbs energy (and therefore chemical potential µ) increases with increasing pressure
Why is the slope of chemical potential (µ) versus temperature steeper for gas than liquid?
Molar entropy of gas is greater than liquid and the slope equals negative molar entropy
(dG/dT)P = -Sm
Why does increasing pressure lower the melting point of ice but raise the melting point of most substances?
According to the Clapeyron equation; dP/dT = ∆S/∆V;
For melting, ∆S > 0 so the sign of slope dP/dT is determined by ∆Vm = Vm(liquid) - Vm(solid)
For most substances ∆Vm > 0 meaning dP/dT is positive, therefore pressure increase raises the melting Temperature;
For water, ∆Vm < 0 causing the slope to be negative and leading to a lower melting temperature with increasing pressure
What assumption is required when using the integrated Clausius-Clapeyron equation to calculate vapor pressure?
The molar enthalpy of vaporization (∆vapH) is assumed constant over the specified temperature interval;
dP/dT = ∆trsH/Ttrs∆trsVm
Why does chemical potential equality define phase equilibrium between two coexisting phases?
Because chemical potential tells us the tendency of a substance to move from one phase to another.
Equal chemical potential (µ1 = µ2) means no net thermodynamic driving force exists for matter transfer between the coexisting phases
What is the difference between a single phase solution and a dispersion?
A single phase solution is uniform at the molecular level (eg. sugar water; solute completely dissolved in a liquid solvent) while a dispersion contains microscopic regions of distinct phases (eg. emulsion of milk reveals 2 phases; oil-in-water)
Why does the triple point of a pure substance have zero degrees of freedom?
Three coexisting phases for one component fix temperature and pressure uniquely so neither variable can be changed without losing a phase
f = c - p + 2
In what case can ∆G not be used to predict whether the process is spontaneous?
When T or P vary; ∆G < 0 is a criterion of spontaneity only if both T and P are held constant.