Chapter 6.3 - Extensive and Intensive Properties

In the previous section, we saw the basic details about enthalpy. We saw some solved examples also. In this section, we will see two more solved examples. Later in this section, we will see extensive and intensive properties.

Solved example 6.9
The combustion of graphite to produce carbon dioxide is represented by the equation C (s) + O₂(g) → CO₂(g). At 298 K and 1.0 atm, ΔH = −393.5 kJ mol^_-1 of graphite for this reaction. What is ΔU for the reaction?
Solution:
• We can solve this problem using two methods
    ♦ Method I is by using the basics.
    ♦ Method II is by using the equations that we derived.
Method I:
1. Let A be the initial state.
• In this state, the system is:
    ♦ 1 mol of carbon at 298 K and 1 atm pressure.
    ♦ 1 mol of O₂ at 298 K and 1 atm pressure.
2. Let B be the final state.
• In this state, the system is:
    ♦ 1 mol of CO₂ at 298 K and 1 atm pressure.
3. We have seen that:
    ♦ Enthalpy in the initial state, H_A = (U_A + PV_A)
    ♦ Enthalpy in the final state, H_B = (U_B + PV_B)
    ♦ So enthalpy change = (H_B - H_A) = [(U_B - U_A) + P(V_B - V_A)]
4. But enthalpy change is given as -393.5 KJ
• So we can write:
[(U_B - U_A) + P(V_B - V_A)] = -393.5
5. In the above equation, there are 3 terms: (U_B - UA), P(V_B - V_A) and -393.5
• So, if we can find P(V_B - V_A), we can calculate (U_B - U_A)
6. P(V_B - V_A) can be calculated in steps:
(i) From ideal gas equation, we have:
    ♦ P_AV_A = n_ART_A
    ♦ P_BV_B = n_BRT_B
• In our present case:
    ♦ P_A = P_B = P = 1 atm
    ♦ T_A = T_B = T = 298 K
(ii) So P(V_B - V_A) = (n_B - n_A)RT
    ♦ n_A = number of gaseous moles in the initial state = number of gaseous moles of O₂ = 1
    ♦ n_B = number of gaseous moles in the final state = number of gaseous moles of CO₂ = 1
    ♦ So (n_B - n_A) = 0
(iii) Substituting the known values, the right side of (ii) becomes:
zero
7. We can use this zero in the place of P(V_B - V_A) in (4)
• We get: [(U_B - U_A) + 0] = -393.5
• So (U_B - U_A) = ΔU = -393.5 kJ

Method II:
1. The pressure is 1 atm (1.013 × 10_⁵ N m_^-2) and temperature is 298 K.
    ♦ Under these conditions, 1 mol of graphite changes into 1 mol CO₂
    ♦ During this process, the pressure and temperature remains the same.
2. Given that, the enthalpy change for this process is -393.5 kJ
    ♦ So ΔH = -393.5 kJ
3. We are asked to find the internal energy change ΔU
    ♦ We have Eq.6.5 which connects ΔH and ΔU: ΔH = ΔU + PΔV
4. So our next aim is to find PΔV
    ♦ We have Eq.6.6: PΔV = Δn_gRT
5. Δn_g = number of gaseous moles in the final state - number of gaseous moles in the initial state.
    ♦ number of gaseous moles in the final state = number of gaseous moles of CO₂ = 1
    ♦ number of gaseous moles in the initial state = number of gaseous moles of O₂ = 1
• So n_g = (1-1) = 0
6. So P ΔV = 0
7. So from (3), we get:
ΔU = (ΔH - PΔV) = (-393.5 - 0) = -393.5 kJ

Solved example 6.10
Calculate ΔU for the conversion of oxygen gas to ozone at 298 K and constant pressure P. 3O₂(g) → 2O₃(g). The value of ΔH for the reaction is 285.4 kJ.
Solution:
• We can solve this problem using two methods.
    ♦ Method I is by using the basics.
    ♦ Method II is by using the equations that we derived Method I.
Method I:
1. Let A be the initial state.
• In this state, the system is:
    ♦ 3 mol of O₂ at 298 K and P atm pressure.
2. Let B be the final state.
• In this state, the system is:
    ♦ 2 mol of O₃ at 298 K and P atm pressure.
3. We have seen that:
    ♦ Enthalpy in the initial state, H_A = (U_A + PV_A)
    ♦ Enthalpy in the final state, H_B = (U_B + PV_B)
    ♦ So enthalpy change = (H_B - H_A) = [(U_B - U_A) + P(V_B - V_A)]
4. But enthalpy change is given as 285.4 KJ
• So we can write:
[(U_B - U_A) + P(V_B - V_A)] = 285.4
5. In the above equation, there are 3 terms: (U_B - UA), P(V_B - V_A) and 285.4
• So, if we can find P(V_B - V_A), we can calculate (U_B - U_A)
6. P(V_B - V_A) can be calculated in steps:
(i) From ideal gas equation, we have:
    ♦ P_AV_A = n_ART_A
    ♦ P_BV_B = n_BRT_B
• In our present case:
    ♦ P_A = P_B = P
    ♦ T_A = T_B = T = 298 K
(ii) So P(V_B - V_A) = (n_B - n_A)RT
    ♦ n_A = number of gaseous moles in the initial state = number of gaseous moles of O₂ = 3
    ♦ n_B = number of gaseous moles in the final state = number of gaseous moles of O₃ = 2
    ♦ So (n_B - n_A) = (3 - 2) = 1
(iii) Substituting the known values, the right side of (ii) becomes:
(1 mol) × (8.3 J mol^_-1 K^_-1) × (298 K) = -2473.4 J = -2.473 kJ
7. We can use the value obtained above, in the place of P(V_B - V_A) in (4)
• We get: [(U_B - U_A) -2.473] = 285.4
• So (U_B - U_A) = ΔU = (285.4 + 2.473) = 287.87 kJ 

Method II:
1. The pressure is P atm and temperature is 298 K
    ♦ Under these conditions, 3 mol of O₂ changes into 2 mol of O₃
    ♦ During this process, the pressure and temperatur remains the same.
2. Given that, the ΔH for this process is 285.4 kJ
3. We are asked to find the internal energy change ΔU
    ♦ We have Eq.6.5 which connects ΔH and ΔU: ΔH = ΔU + PΔV
4. So our next aim is to find PΔV
    ♦ We have Eq.6.6: PΔV = Δn_gRT
5. Δn_g = number of gaseous moles in the final state - number of gaseous moles in the initial state
    ♦ number of gaseous moles in the final state = number of gaseous moles of O₃ = 2
    ♦ number of gaseous moles in the initial state = number of gaseous moles of O₂ = 3
• So n_g = (2-3) = -1
6. Substituting the known values in (4), we get:
PΔV = (-1 mol) × (8.3 J mol^_-1 K^_-1) × (298 K) = -2473.4 J = -2.473 kJ
7. So from (3), we get:
ΔU = (ΔH - PΔV) = [285.4 - (-2.473)] = 287.87 kJ

Solved example 6.11
ΔU of combustion of methane is – X kJ mol^-1 . The value of ΔH is
(i) = ΔU
(ii) > ΔU
(iii) < ΔU
(iv) = 0
Solution:
• We will solve this problem using Method I only.
• The reader may try Method II also.
Method I:
1. The balanced equation for the combustion of methane is:
CH₄(g) + 2O₂(g)  → CO₂(g) + 2H₂O(l)
• Let A be the initial state
• In this state, the system is:
    ♦ 1 mol of CH₄ and 2 mol O₂ at 298 K and P atm pressure
2. Let B be the final state
• In this state, the system is:
    ♦ 1 mol of CO₂ and 2 mol H₂O at 298 K and P atm pressure
3. We have seen that:
    ♦ Enthalpy in the initial state, H_A = (U_A + PV_A)
    ♦ Enthalpy in the final state, H_B = (U_B + PV_B)
    ♦ So enthalpy change = (H_B - H_A) = [(U_B - U_A) + P(V_B - V_A)]
4. Here, (U_B - U_A) is given as -X KJ
• So we can write:
[(-X) + P(V_B - V_A)] = H_B - H_A
5. In the above equation, there are 3 terms: (-X), P(V_B - V_A) and (H_B - H_A)
• So, if we can find P(V_B - V_A), we can calculate (H_B - H_A)
6. P(V_B - V_A) can be calculated in steps:
(i) From ideal gas equation, we have:
    ♦ P_AV_A = n_ART_A
    ♦ P_BV_B = n_BRT_B
• In our present case:
    ♦ P_A = P_B = P
    ♦ T_A = T_B = T = 298 K
(ii) So P(V_B - V_A) = (n_B - n_A)RT
    ♦ n_A = number of gaseous moles in the initial state = number of gaseous moles of CH4 and O₂ = (1+2) = 3
    ♦ n_B = number of gaseous moles in the final state = number of gaseous moles of CO₂ = 1
    ♦ So (n_B - n_A) = (1 - 3) = -2
(iii) Substituting the known values, the right side of (ii) becomes:
(-2 mol) × (8.3 J mol_^-1 K_^-1) × (298 K) = -4946.8 J = -4.946 kJ
7. We can use the value obtained above, in the place of P(V_B - V_A) in (4)
• We get: [(-X) -4.946] = (H_B - H_A)
• So (H_B - H_A) = ΔH = -[X+4.946]
8. -[X+4.946] is less than -X
• So the correct option is (iii)

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Extensive and intensive properties can be explained in 6 steps:
1. In fig.6.7 below, the outlines of a container is shown in yellow color.

Fig.6.7

• Inside the container, there is a gas at temperature T.
• Volume of the container is V. So volume of the gas is also V.
2. In fig.b, the container is divided into two equal parts by the red partition.
• The volume of gas in each half of the container will be V/2.
• The temperature of the gas in each half of the container will be the same T.
    ♦ We can insert a thermometer at any point in figs. (a) and (b)
    ♦ All readings will be T.
3. Let m be the mass of the gas in fig.a
• Then in fig.b, the mass in each half will be m/2
• We can take different samples from different parts in figs.(a) and (b)
    ♦ The mass of all those samples will be different from m.
4. Let d be the density of the gas in fig.a
• Then in fig.b, density in each half will be the same d.
• We can take different samples from different parts in fig.(a) and (b)
    ♦ The density of all those samples will be the same d.
(Recall that, the density of a small piece of iron will be same as that of a large block of iron)
5. We see that:
• Volume and mass depend on the size of the container.
• Temperature and density do not depend on the size of the container.
■ There are some properties whose values depend on quantity and size of matter present in the system. Such properties are called extensive properties.
    ♦ Volume and mass are extensive properties.
■ There are some properties whose values do not depend on quantity and size of matter present in the system. Such properties are called intensive properties.
    ♦ Temperature and density are intensive properties.
6. Let us see some more examples for extensive and intensive properties:
(i) Internal energy U is an extensive property.
• Recall that, U is the sum of kinetic and potential energies of all the molecules in the system.
• So, if the quantity of matter in the system changes, the number of molecules will change and so, U will change.
(ii) Enthalpy H is an extensive property.
• We have: H = U + P ΔV
• U is an extensive property.
• ΔV is change in volume. It is an extensive property
• So H will be an extensive property.
(iii) Pressure P is an intensive property.
• We can insert a pressure gauge at any point in the system. The readings will be the same.
• Imagine that, we take out a small sample from the system.
    ♦ The number of molecules in that sample will be less.
    ♦ So we are inclined to think that, pressure in that sample will be lower.
• But remember that, pressure is inversely proportional to volume.
    ♦ The volume of the sample that we take out is lesser.
    ♦ So the pressure will increase according to the inverse proportion.

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• In addition to extensive property and intensive property, there is one more item. It is called molar property
• Molar property can be explained in 4 steps:
1. Consider any extensive property. Let us denote it as χ
2. Since it is an extensive property, it will depend on the size or quantity of matter.
• For example, let χ be the volume V
• Then we can write:
    ♦ χ for 1 kg = χ_{(1 kg)} = V_{(1 kg)}
    ♦ χ for 2 kg = χ_{(2 kg)} = V_{(2 kg)}
• V_{(1 kg)} will be different from V_{(2 kg)} because, V depends on the size and amount of matter.
3. All extensive properties depend on the size and amount of matter.
• In this situation, we define a new property called molar property.
■ It is the value of the extensive property of the system for one mol of the substance.
• It is denoted as: χ_m
• For example, if χ is the extensive property V, the volume of one mol of the substance will be denoted as V_m
4. χ_m can be obtained by dividing the value of χ by the number of moles present in the system.
• Thus we get Eq.6.7: $\mathbf\small{\rm{\chi_m=\frac{\chi}{n}}}$
• For example, if V is the volume of n moles of a substance, the molar volume V_m can be obtained as: $\mathbf\small{\rm{V_m=\frac{V}{n}}}$

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Heat capacity
• Detailed notes on heat capacity can be seen in chapter 11.3 of Physics lessons
• Based on those notes, we get information about:
Heat capacity, Specific heat capacity, Molar specific heat capacity, Specific heat capacity at constant pressure (C_P) and Specific heat capacity at constant volume (C_V).
• Also we get the method to find the amount of heat absorbed by a body.

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Now we can find the relation between C_P and C_V for an ideal gas. It can be written in 10 steps:
1. Consider a system consisting of an ideal gas. Let us denote the 'heat capacity of the gas' by the letter ‘C’
• Then we can write:
    ♦ C_P is the heat capacity at constant pressure.
    ♦ C_V is the heat capacity at constant volume.
(Note that, 'C' denote the heat capacity. Not the specific heat capacity. ‘Specific heat capacity’ is related to unit mass of the substance. When we use ‘heat capacity’, the total mass of the substance is automatically included)
2. First we examine the constant volume condition.
• Let the system be heated at constant volume.
• Let the change in temperature be ΔT.
■ Then the heat absorbed by the gas will be equal to C_V ΔT
• Since this heat is absorbed at constant volume, we can denote it as Q_V
• So we can write: Q_V = C_V ΔT
3. Since volume is constant, no work is done by the gas. So all the heat absorbed will be used to increase the internal energy of the system.
• We can write: Q_V = ΔU = C_V ΔT
4. Now we examine the constant pressure condition.
• Let the system be heated at constant pressure (P) in such a way that, the rise in temperature is the same ΔT mentioned in (2)
• Then the heat absorbed by the system will be equal to C_P ΔT
5. Since this heat is absorbed at constant pressure, we can denote it as Q_P
• So we can write: Q_P = C_P ΔT
6. Since the volume is not constant, work is done by the system. So the absorbed heat is used for two items:
    ♦ Increase the internal energy.
    ♦ Doing work against the external pressure
• We can write: Q_P = ΔU + P ΔV
7. To be more clear, we will write the result in (6) in a basic form:
Q_P = (U_B – U_A) + P (V_B – V_A)
⇒ Q_P = (U_B – U_A) + PV_B – PV_A
8. We have the ideal gas equation: PV = nRT
• So PV_B = n_BRT_B and PV_A = n_ART_A
• So the result in (7) becomes:
Q_P = (U_B – U_A) + n_BRT_B – n_ART_A
⇒ Q_P = (U_B – U_A) + R(n_BT_B – n_AT_A)
9. Let us assume that, the system has only one mole of the gas. Also there is no leakage of gas.
• Then we get: n_B = n_A = 1
• So the result in (8) becomes:
Q_P = (U_B – U_A) + R(T_B – T_A)
⇒ Q_P = (U_B – U_A) + R ΔT
⇒ Q_P = ΔU + R ΔT
10. But from (5), we have: Q_P = C_P ΔT
• Also from (3), we have: ΔU = Q_V = C_V ΔT
• So the result in (9) becomes:
C_P ΔT = C_V ΔT + R ΔT
⇒ C_P = C_V + R
• Thus we get Eq.6.8: C_P - C_V = R
■ This is called Mayer's relation

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In the next section we will see calorimetry


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