Energetics & Equilibria 1: first law, second law, gas expansions, internal energy, enthalpy, heat capacity, absolute entropy Flashcards
2nd law of thermodynamics
In a spontaneous process, the entropy of the universe increases
ΔSuniv > 0
What are spontaneous reactions?
Reactions or processes which occur without intervention/facilitation, and where the entropy of the universe increases.
Their reverses are non-spontaneous but can often be forced.
Energy minimisation is not the criterion for spontaneous processes. Not all spontaneous reactions are exothermic - e.g. an equilibrium can be approached from both sides.
units of entropy
J K-1 or J K-1 mol-1
For a system with a fixed number of molecules, fixed volume and fixed amount of energy, with no exchange of molecules or energy:
- A configuration is one possible arrangement of molecules amongst the energy levels
- For a given configuration, different molecules (but always the same population, i.e. number of molecules) may be found in a given energy level
- The weight of a configuration, W, is the number of ways of achieving it
give the formula for W
W = N! / n0! x n1! x n2! x … x nn!
Where N is the total number of molecules and nn is the population of one energy level
what is the Boltzmann distribution?
The Boltzmann (most probable) distribution is the configuration with the largest weight (significantly larger when the number of molecules is large).
formula for Boltzmann distribution
ni = n0e−εi /kT
- ni is the number of molecules in energy level i
- εi is the energy of i
- n0 is the population of the ground level, which has energy 0
- k is Boltzmann’s constant (k = R/NA, where NA is Avogadro’s constant)
Verbally, as the energy of a level increases, its population decreases. Levels with energies less than or similar to kT have significant populations. So ground level is most populated
Entropy is related to the number of ways of achieving a configuration. Give its formula in terms of this.
S = k lnW
k = Boltzmann constant
An object at temperature T reversibly absorbs a small amount of heat. Give the formula for the entropy change.
dS = δqrev/T
where dδqrev = small amount of heat
NB: If we have an irreversible process taking us from A to B, then, in order to calculate the entropy change, we need to work out what the heat would be if we were to go from A to B by a reversible path.
Explain why the following formula make the same prediction of the effect on entropy of increasing heat:
- S = k lnW
- dS = δqrev/T
- Heat a system → molecules’ internal energy increases → they move to higher energy levels → more ways W of achieving distribution → lnW increases → S, entropy, increases
- Absorb heat → δqrev is positive (endothermic) → dS is positive, so entropy increases
Give the formula for the entropy change of the universe
∆Suniv = ∆Ssys + ∆Ssurr
= ∆Ssys - qsys/Tsys
Explain how to get from the first to second eq:
∆Suniv = ∆Ssys + ∆Ssurr
∆Suniv = ∆Ssys - qsys/Tsys
System is closed (no matter is exchanged between system and surroundings), so ∆Ssurr is purely the result of heat exchange.
Surroundings are large, so:
- Tsurr is constant
- Heat exchange with surroundings is reversible from the POV of the surroundings*, so dS = δqrev/T applies
So ∆Ssurr = -qsurr/Tsurr
But
- qsurr = -qsys
- Tsurr = Tsys
So: ∆Suniv = ∆Ssys + (-qsys/Tsys)
= ∆Ssys - qsys/Tsys
*reversible since surroundings are unaffected by heat, so an infinitesimal change (eg in temp) will cause the direction to change
Why does water have to be made cold to freeze?
Water has to be made cold to freeze, so that Tsurr (= Tsys) is negative enough to make ∆Ssurr larger than ∆Ssys, and thus make ∆Suniv positive overall. This is required since 2nd law states that entropy of universe increases in a spontaneous process.
∆Suniv = ∆Ssys - qsys/Tsys
- state the first law of thermodynamics
- give the equation for it
Energy cannot be created or destroyed, only converted from one form to another:
∆U = q + w
Define all the terms in this equation: ∆U = q + w
∆U is internal energy (a property: the energy stored in a substance, in particles’ kinetic energy and energy of interaction; affected by heat and work)
q is the heat absorbed by a system (heat is the means by which energy transfers from hotter bodies to cooler ones to equalise their temperatures. Is positive when a system absorbs heat)
w is the work done on the system (work is a form of energy; w = Fx (force x distance). Is positive when work is done on a system)
what can be said about the internal energy of an ideal gas?
For an ideal gas, all internal energy is in the form of particles’ kinetic energy (since there are no interactions between particles).
Compare state and path functions.
State functions have the same values for a given state (no matter how the state was approached). e.g. temp, pressure, moles, etc
The value of a path function depends on the path a system follows when switching between states.
∆U = q + w
Which terms are state functions and which are path functions?
U is a state function
q and w are path functions
since, as long as ∆U = q + w, the value of U is the same, no matter which values q and w take.
w is the work done on a system. what symbol is given to the work done by a system?
w’, where w’ = -w
a process is exothermic. is q +ve or -ve?
q = heat absorbed by the system
exothermic means heat is given out, so q is -ve
why is it incorrect to write ∆q?
∆q implies that ∆q = qA - qB, where A and B are certain states at which q has a fixed value. But q is a path function, so this isn’t the case.
so you only ever write q. whereas ∆ symbol is applicable for state functions
ideal gas equation and units used?
pV = nRT
- P in Pa (Nm-1)
- V in m3
- n in mol
- T in K
- R = 8.3145 J K-1 mol-1
when is it reasonable to assume the ideal gas law applies?
- for an ideal gas, all internal energy is kinetic energy, i.e. particles don’t interact significantly
- can assume this for modest temperatures and pressures
A gas is confined inside a cylinder by a piston. Give the expression for the work done by the gas when the piston moves.
δw’ = pextdV
work done by gas = external pressure x change in volume of gas
A gas is confined inside a cylinder by a piston.
Derive that δw = -pextdV, stating any assumptions made.
- recall w’ = work done by system*
- pext = external pressure, i.e. pressure caused by external gas*
Assume that the piston is massless and that it moves without friction, so that we only need to consider work done by gas.
Work done = force x distance
Use calculus language: small amount of work done, δw’, by gas, against a force, causes the piston to move by small distance dx:
δw’ = force x dx
The force the system moves against is due to the external gas, and force = pext x area, so:
δw’ = pextA x dx
But area x distance = volume, so Adx = dV
So δw’ = pextdV
- recall w’ = -w, so:*
- δw = pextdV
So δw = -pextdV
expansion against constant external pressure
Integrate the relationship for work done by a gas, when the gas expands from volume vi to vf, against constant external pressure.
State any assumptions made.
Recall δw’ = pextdV
Define reversible and irreversible processes.
Reversible processes: can be changed in direction by an infinitessimal change in some variable:
- Work and heat are at maxima
- Infinitely slow
- At equilibrium
Irreversible processes: cannot be changed in direction by an infinitessimal change in some variable:
- Spontaneous
- Do less than maximum work
- Finite rate
- Not at equilibrium
Explain why, since work done in a reversible process is a maximum, heat absorbed must also be a maximum.
- ∆U = q + w = q - w’ (first law), so q = ∆U + w’
- So, if the path changes such that w’ is maximised, q must change by the same amount in order to keep ∆U contant (which must be true since state function)
- So q must also be maximum
What is the condition for a gas expansion doing maximum work?
Recall δw’ = pextdV
External pressure should be infinitessimally smaller than the internal pressure (since want pext to be as large as possible to maximise δw’, but not so large that it exceeds internal pressure, and thus the gas is compressed rather than expanded).
What does isothermal mean?
Constant temperature
e.g. could be achieved by surrounding gas expansion system with a regulated water bath
reversible isothermal expansion of a gas
Integrate the relationship for work done by a gas, when the gas expands isothermally from volume vi to vi.
State any assumptions made.
Assume
- gas is ideal
- expansion is reversible, so pext = pint (effectively)
- constant temp (isothermal)
What expression gives the heat absorbed in a reversible, isothermal expansion of an ideal gas, from volume Vi to Vf?
qrev = nRTln(Vf/Vi)
Explain why, for the reversible, isothermal expansion of an ideal gas from volume Vi to Vf:
qrev = nRTln(Vf/Vi)
- For an ideal gas, internal energy U consists entirely of kinetic energy, i.e. it only depends on temperature
- Therefore, at constant temperature (isothermal), altering pressure or volume doesn’t affect U
- So ∆U = 0
- ∆U = q + w, so q + w = 0, so q = -w
- recall -w = w’, so w’ = q
- recall w’ = nRTln(Vf/Vi) (flashcard 30)
- so q = nRTln(Vf/Vi)
What expression gives the entropy change of the gas in a reversible, isothermal expansion of an ideal gas, from volume Vi to Vf?
∆S = nRln(Vf/Vi)
For a reversible, isothermal expansion of an ideal gas, from volume Vi to Vf, derive that:
∆S = nRln(Vf/Vi)
- For an ideal gas, internal energy U consists entirely of kinetic energy, i.e. it only depends on temperature
- Therefore, at constant temperature (isothermal), altering pressure or volume doesn’t affect U
- So ∆U = 0
- ∆U = q + w, so q + w = 0, so q = -w
- recall -w = w’, so w’ = q
- recall w’ = nRTln(Vf/Vi) (flashcard 30)
- so q = nRTln(Vf/Vi)
- ∆S = δqrev/T = nRTln(Vf/Vi)/T
- So ∆S = nRln(Vf/Vi)
What is the heat equal to under:
- constant volume conditions?
- constant pressure conditions?
- ∆U, change in internal energy
- ∆H, enthalpy change
- Write ∆U = q + w in differential form
- Assume the only work done is due to expansion of an ideal gas. Alter the equation
- dU = δq + δw
* Use δq and δw rather than dq and dw, since “d” implies “a small change”, but they are state functions, so δ is used to imply “a small amount”*
2. dU = δq - pdV (pV work only)
Recall δw’ = pdV so δw = - pdV
When is it true that dU = δq?
In a sealed container at constant volume, the gas can’t expand, so no work can be done, so pdV = 0, so heat absorbed = change in internal energy.
dU = δq (const. vol)
Give the definition of the constant volume heat capacity.
Derive that:
As more heat is supplied, temperature increases:
q = CV∆T = nCV,m∆T
- Where Cm = molar heat capacity and n = amount in moles*
- ∆U = q + w = q - w’ = q - pdV*
At constant volume, pdV = 0, so ∆U = q, so:
∆U = nCV,m∆T (constant volume)
∆Um = CV,m∆T (constant volume)Where CV,m = molar heat capacity at constant volume and Um = change in molar internal energy
Write in differential form:
dUm = CV,mdT (constant volume)
Rearrange:
CV,m = dUm/dT (constant volume)
- Express as a partial derivative, where ∂ indicates only variation of U wrt T is being considered, and V indicates constant volume:*
- CV,m = (∂Um/∂T)V*
When is it true that dH = δq?
Heat is equal to enthalpy change under constant pressure conditions
State the definition of enthalpy (eq)
H = U + pV
Derive that dH = δq (constant pressure)
Definition of enthalpy
H = U + pV
Differentiate both sides:
dH = dU + d(pV)
Product rule for d(pV)
dH = dU + pdV + Vdp
Recall first law dU = δq + δw = δq - pdV
dH = δq - pdV + pdV + Vdp
dH = δq + Vdp
At constant external pressure, Vdp = 0, so:
dH = δq (const. pressure)
Give the definition of the molar heat capacity at constant pressure
Cp,m = (∂Hm/∂T)p
Derive that
Cp,m = (∂Hm/∂T)p
As more heat is supplied, temperature increases:
q = C∆T = nCm∆T
Where Cm = molar heat capacity and n = amount in moles
At constant pressure, ∆H = q, so:
∆Hm = Cp,m∆T (constant pressure)
- Where Cp,m = molar heat capacity at constant pressure and* ∆Hm = change in molar enthalpy
- Write in differential form:*
dHm = Cp,mdT (constant pressure)
Rearrange:
Cp,m = dHm/dT (constant pressure)
Express as a partial derivative, where ∂ indicates only variation of H wrt T is being considered, and p indicates constant pressure:
Cp,m = (∂Hm/∂T)p
The molar enthalpy of a substance is known at temperature T1, Hm(T1). Derive an expression for the molar enthalpy at temperature T2, Hm(T2).
Definition of heat capacity at constant pressure:
Cp,m = (∂Hm/∂T)p
Cp,m = dHm/dT (constant pressure)
dHm = Cp,mdT (constant pressure)
- Then follow the image*
- NB we can only measure enthalpy changes rather than absolute values. A later fc will modify this eq to convert ∆H values between different temps*
How is the heat capacity at a given temp
- measured?
- evaluated?
- Measure temperature rise of substance for a given heat input. I.e. for a plot of heat input against temp, heat capacity at a particular temp = gradient of tangent
- Heat capacity varies weakly with temp, so we work around it by expressing it in parametrised function form C(T) = a + bT + d/T2. Values of a/b/c are tabulated, so heat capacity can be evaluated at any temp
Show that S = 0 at absolute zero.
At absolute 0, kT –> 0
So all molecules are in the ground state
So there is only one way of arranging them: W = 1
S = klnW = k x 0 = 0
How are the molar entropy and enthalpy associated with a phase change related?
ΔSm,pc = ΔHm,pc/Tpc
Where ΔSm,pc = molar entropy associated with a phase change, and ΔHm,pc = molar enthalpy associated with a phase change
Absolute entropies can be tabulated/evaluated using heat capacity measurements.
Say the heat capacity is measured from temperature T = 0 to T = T*. Derive how the entropy at T* would be found, stating any assumptions made.
Defintion of entropy:
dS = δqrev/T
At constant pressure, heat = enthalpy change, so:
dS = dH/T (constant pressure)
Switch to molar quantities:
dSm = dHm//T
Definition of constant pressure heat capacity:
Cp,m = (∂Hm/∂T)p
Cp,m = dHm/dT (constant pressure)
dHm = Cp,mdT (constant pressure)
Sub the definition of heat capacity into that of entropy:
dSm = Cp,mdT/T (constant pressure)
To find entropy at temperature T*, integrate both sides from T = 0 to T = T*: (top image)
Assumption: Cp,m(T) is constant over a small temp range
Measure Cp,m for different temps, then plot Cp,m(T)/T against temperature, from 0 to T*
Area under curve = Sm(T*) - Sm(0)
At absolute 0, entropy = 0, so area = Sm(T*)
Account for phase changes which occur over the temp range measured:
ΔSm,pc = ΔHm,pc/Tpc
Where ΔSm,pc = molar entropy associated with a phase change, and ΔHm,pc = molar enthalpy associated with a phase change
bottom image = final answer
Derive a relationship for converting molar entropies from one temperature to another.
fc 49:
Recall dSm = Cp,mdT/T (constant pressure)
Integrate between T1 and T2 (image):
assumption: temperatures are not very different, so Cp is constant and can be taken out of the integral
