Energetics & Equilibria 1: first law, second law, gas expansions, internal energy, enthalpy, heat capacity, absolute entropy

Question 1

2nd law of thermodynamics

Answer 1

In a spontaneous process, the entropy of the universe increases

ΔS_univ > 0

Question 2

What are spontaneous reactions?

Answer 2

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.

Question 3

units of entropy

Answer 3

J K-1 or J K-1 mol-1

Question 4

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

Answer 4

W = N! / n₀! x n₁! x n₂! x … x n_n!

Where N is the total number of molecules and n_n is the population of one energy level

Question 5

what is the Boltzmann distribution?

Answer 5

The Boltzmann (most probable) distribution is the configuration with the largest weight (significantly larger when the number of molecules is large).

Question 6

formula for Boltzmann distribution

Answer 6

n_i = n₀e^{−εi /kT}

• n_i is the number of molecules in energy level i
• ε_i is the energy of i
• n₀ is the population of the ground level, which has energy 0
• k is Boltzmann’s constant (k = R/N_A, where N_A 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

Question 7

Entropy is related to the number of ways of achieving a configuration. Give its formula in terms of this.

Answer 7

S = k lnW

k = Boltzmann constant

Question 8

An object at temperature T reversibly absorbs a small amount of heat. Give the formula for the entropy change.

Answer 8

dS = δq_rev/T

where dδq_rev = 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.

Question 9

Explain why the following formula make the same prediction of the effect on entropy of increasing heat:

1. S = k lnW
2. dS = δq_rev/T

Answer 9

1. Heat a system → molecules’ internal energy increases → they move to higher energy levels → more ways W of achieving distribution → lnW increases → S, entropy, increases
2. Absorb heat → δq_rev is positive (endothermic) → dS is positive, so entropy increases

Question 10

Give the formula for the entropy change of the universe

Answer 10

∆S_univ = ∆S_sys + ∆S_surr

= ∆S_sys - q_sys/T_sys

Question 11

Explain how to get from the first to second eq:

∆S_univ = ∆S_sys + ∆S_surr

∆S_univ = ∆S_sys - q_sys/T_sys

Answer 11

System is closed (no matter is exchanged between system and surroundings), so ∆S_surr is purely the result of heat exchange.

Surroundings are large, so:

• T_surr is constant
• Heat exchange with surroundings is reversible from the POV of the surroundings*, so dS = δq_rev/T applies

So ∆S_surr = -q_surr/T_surr

But

• q_surr = -q_sys
• T_surr = T_sys

So: ∆S_univ = ∆S_sys + (-q_sys/T_sys)

= ∆S_sys - q_sys/T_sys

*reversible since surroundings are unaffected by heat, so an infinitesimal change (eg in temp) will cause the direction to change

Question 12

Why does water have to be made cold to freeze?

Answer 12

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

Question 13

• state the first law of thermodynamics
• give the equation for it

Answer 13

Energy cannot be created or destroyed, only converted from one form to another:

∆U = q + w

Question 14

Define all the terms in this equation: ∆U = q + w

Answer 14

∆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)

Question 15

what can be said about the internal energy of an ideal gas?

Answer 15

For an ideal gas, all internal energy is in the form of particles’ kinetic energy (since there are no interactions between particles).

Question 16

Compare state and path functions.

Answer 16

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.

Question 17

∆U = q + w

Which terms are state functions and which are path functions?

Answer 17

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.

Question 18

w is the work done on a system. what symbol is given to the work done by a system?

Answer 18

w’, where w’ = -w

Question 19

a process is exothermic. is q +ve or -ve?

Answer 19

q = heat absorbed by the system

exothermic means heat is given out, so q is -ve

Question 20

why is it incorrect to write ∆q?

Answer 20

∆q implies that ∆q = q_A - q_B, 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

Question 21

ideal gas equation and units used?

Answer 21

pV = nRT

• P in Pa (Nm^-1)
• V in m³
• n in mol
• T in K
• R = 8.3145 J K^-1 mol^-1

Question 22

when is it reasonable to assume the ideal gas law applies?

Answer 22

• 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

Question 23

A gas is confined inside a cylinder by a piston. Give the expression for the work done by the gas when the piston moves.

Answer 23

δw’ = p_extdV

work done by gas = external pressure x change in volume of gas

Question 24

A gas is confined inside a cylinder by a piston.

Derive that δw = -p_extdV, stating any assumptions made.

Answer 24

• recall w’ = work done by system*
• p_ext = 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 = p_ext x area, so:

δw’ = p_extA x dx

But area x distance = volume, so Adx = dV

So δw’ = p_extdV

• recall w’ = -w, so:*
• δw = p_extdV

So δw = -p_extdV

Question 25

expansion against constant external pressure

Integrate the relationship for work done by a gas, when the gas expands from volume v_i to v_f, against constant external pressure.

State any assumptions made.

Answer 25

Recall δw’ = p_extdV

Question 26

Define reversible and irreversible processes.

Answer 26

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

Question 27

Explain why, since work done in a reversible process is a maximum, heat absorbed must also be a maximum.

Answer 27

• ∆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

Question 28

What is the condition for a gas expansion doing maximum work?

Answer 28

Recall δw’ = p_extdV

External pressure should be infinitessimally smaller than the internal pressure (since want p_ext 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).

Question 29

What does isothermal mean?

Answer 29

Constant temperature

e.g. could be achieved by surrounding gas expansion system with a regulated water bath

Question 30

reversible isothermal expansion of a gas

Integrate the relationship for work done by a gas, when the gas expands isothermally from volume v_i to v_i.

State any assumptions made.

Answer 30

Assume

• gas is ideal
• expansion is reversible, so p_ext = p_int (effectively)
• constant temp (isothermal)

Question 31

What expression gives the heat absorbed in a reversible, isothermal expansion of an ideal gas, from volume V_i to V_f?

Answer 31

q_rev = nRTln(V_f/V_i)

Question 32

Explain why, for the reversible, isothermal expansion of an ideal gas from volume V_i to V_f:

q_rev = nRTln(V_f/V_i)

Answer 32

• 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(V_f/V_i) (flashcard 30)
• so q = nRTln(V_f/V_i)

Question 33

What expression gives the entropy change of the gas in a reversible, isothermal expansion of an ideal gas, from volume Vi to Vf?

Answer 33

∆S = nRln(V_f/V_i)

Question 34

For a reversible, isothermal expansion of an ideal gas, from volume Vi to Vf, derive that:

∆S = nRln(V_f/V_i)

Answer 34

• 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 = δq_rev/T = nRTln(Vf/Vi)/T
• So ∆S = nRln(V_f/V_i)

Question 35

What is the heat equal to under:

• constant volume conditions?
• constant pressure conditions?

Answer 35

• ∆U, change in internal energy
• ∆H, enthalpy change

Question 36

1. Write ∆U = q + w in differential form
2. Assume the only work done is due to expansion of an ideal gas. Alter the equation

Answer 36

1. 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

Question 37

When is it true that dU = δq?

Answer 37

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)

Question 38

Give the definition of the constant volume heat capacity.

Answer 38

Question 39

Derive that:

Answer 39

As more heat is supplied, temperature increases:

q = C_V∆T = nC_V,m∆T

• Where C_m = 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 = nC_V,m∆T (constant volume)

∆U_m = C_V,m∆T (constant volume)Where C_V,m = molar heat capacity at constant volume and U_m = change in molar internal energy

Write in differential form:

dU_m = C_V,mdT (constant volume)

Rearrange:

C_V,m = dU_m/dT (constant volume)

• Express as a partial derivative, where ∂ indicates only variation of U wrt T is being considered, and V indicates constant volume:*
• C_V,m = (∂U_m/∂T)_V*

Question 40

When is it true that dH = δq?

Answer 40

Heat is equal to enthalpy change under constant pressure conditions

Question 41

State the definition of enthalpy (eq)

Answer 41

H = U + pV

Question 42

Derive that dH = δq (constant pressure)

Answer 42

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)

Question 43

Give the definition of the molar heat capacity at constant pressure

Answer 43

C_p,m = (∂H_m/∂T)_p

Question 44

Derive that

C_p,m = (∂H_m/∂T)_p

Answer 44

As more heat is supplied, temperature increases:

q = C∆T = nC_m∆T

Where C_m = molar heat capacity and n = amount in moles

At constant pressure, ∆H = q, so:

∆H_m = C_p,m∆T (constant pressure)

• Where C_p,m = molar heat capacity at constant pressure and* ∆H_m = change in molar enthalpy
• Write in differential form:*

dH_m = C_p,mdT (constant pressure)

Rearrange:

C_p,m = dH_m/dT (constant pressure)

Express as a partial derivative, where ∂ indicates only variation of H wrt T is being considered, and p indicates constant pressure:

C_p,m = (∂H_m/∂T)_p

Question 45

The molar enthalpy of a substance is known at temperature T₁, H_m(T₁). Derive an expression for the molar enthalpy at temperature T₂, H_m(T₂).

Answer 45

Definition of heat capacity at constant pressure:

C_p,m = (∂H_m/∂T)_p

C_p,m = dH_m/dT (constant pressure)

dH_m = C_p,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*

Question 46

How is the heat capacity at a given temp

• measured?
• evaluated?

Answer 46

• 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

Question 47

Show that S = 0 at absolute zero.

Answer 47

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

Question 48

How are the molar entropy and enthalpy associated with a phase change related?

Answer 48

ΔS_m,pc = ΔH_m,pc/T_pc

Where ΔS_m,pc = molar entropy associated with a phase change, and ΔH_m,pc = molar enthalpy associated with a phase change

Question 49

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.

Answer 49

Defintion of entropy:

dS = δq_rev/T

At constant pressure, heat = enthalpy change, so:

dS = dH/T (constant pressure)

Switch to molar quantities:

dS_m = dH_m//T

Definition of constant pressure heat capacity:

C_p,m = (∂H_m/∂T)_p

C_p,m = dH_m/dT (constant pressure)

dH_m = C_p,mdT (constant pressure)

Sub the definition of heat capacity into that of entropy:

dS_m = C_p,mdT/T (constant pressure)

To find entropy at temperature T*, integrate both sides from T = 0 to T = T*: (top image)

Assumption: C_p,m(T) is constant over a small temp range

Measure C_p,m for different temps, then plot C_p,m(T)/T against temperature, from 0 to T*

Area under curve = S_m(T*) - S_m(0)

At absolute 0, entropy = 0, so area = S_m(T*)

Account for phase changes which occur over the temp range measured:

ΔS_m,pc = ΔH_m,pc/T_pc

Where ΔS_m,pc = molar entropy associated with a phase change, and ΔH_m,pc = molar enthalpy associated with a phase change

bottom image = final answer

Question 50

Derive a relationship for converting molar entropies from one temperature to another.

Answer 50

fc 49:

Recall dS_m = C_p,mdT/T (constant pressure)

Integrate between T₁ and T₂ (image):

assumption: temperatures are not very different, so C_p is constant and can be taken out of the integral