Please Wait... While Loading Full Video#### Class 11 Chapter 6 Thermodynamics

### Applications-II

● Enthalpy, `H`

● Extensive and Intensive Properties

● Heat Capacity

● The Relationship between `C_p` and `C_V` of an Ideal Gas

● Extensive and Intensive Properties

● Heat Capacity

● The Relationship between `C_p` and `C_V` of an Ideal Gas

`=>` The heat absorbed at constant volume is equal to change in the internal energy i.e., `color{purple}(ΔU = q_V)`.

● But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure. So, we need to define another state function which may be suitable under these conditions.

`=>` We may write equation (6.1) as `color{purple}(ΔU = q_p− pΔV)` at constant pressure, where `q_p` is heat absorbed by the system and `color{purple}(–pΔV)` represent expansion work done by the system.

● Let us represent the initial state by subscript `1` and final state by `2`. We can rewrite the above equation as

`color{purple}(U_2–U_1 = q_p – p (V_2 – V_1)`

● On rearranging, we get

`color{purple}(q_p = (U_2 + pV_2) – (U_1 + pV_1))` ........(6.6)

Now we can define another thermodynamic function, the enthalpy `H` [Greek word enthalpien, to warm or heat content] as :

`color{purple}(H = U + pV)` .................... (6.7)

So, equation (6.6) becomes

`color{purple}(q_p= H_2 – H_1 = ΔH)`

`=>` Although `color{purple}(q)` is a path dependent function, `color{purple}(H)` is a state function because it depends on `U`, `p` and `V`, all of which are state functions.

● Therefore, `color{purple}(ΔH)` is independent of path. Hence, `color{purple}(q_p)` is also independent of path.

`=>` For finite changes at constant pressure, we can write equation 6.7 as

`color{purple}(ΔH = ΔU + ΔpV)`

Since `color{purple}(p)` is constant, we can write

`color{purple}(ΔH = ΔU + pΔV)` .................... (6.8)

`color{red}("Note ")` When heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.

`=>` `color{purple}(ΔH = q_p)`, heat absorbed by the system at constant pressure.

● `color{purple}(ΔH)` is negative for `color{red}("exothermic reactions")` which evolve heat during the reaction

● `color{purple}(ΔH)` is positive for `color{red}("endothermic reactions")` which absorb heat from the surroundings.

`=>` At constant volume `color{purple}((ΔV = 0), ΔU = q_V),` therefore equation 6.8 becomes

`color{purple}(ΔH = ΔU = q_V)`

`=>` The difference between `color{purple}(ΔH)` and `color{purple}(ΔU)` is not usually significant for systems consisting of only solids and/or liquids. Solids and liquids do not suffer any significant volume changes upon heating.

`=>` The difference becomes significant when gases are involved.

● Let us consider a reaction involving gases.

● If `color{purple}(V_A)` is the total volume of the gaseous reactants, `color{purple}(V_B)` is the total volume of the gaseous products, `n_A` is the number of moles of gaseous reactants and `color{purple}(n_B)` is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,

`color{purple}(pV_A = n_ART)` and `color{purple}(pV_B = n_BRT)`

Thus, `color{purple}(pV_B – pV_A = n_BRT – n_ART = (n_B–n_A)RT)`

or `color{purple}(p (V_B – V_A) = (n_B – n_A) RT)`

or` color{purple}(p ΔV = Δn_gRT)` ................. (6.9)

● Here, `color{purple}(Δn_g)` refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.

● Substituting the value of `color{purple}(pΔV)` from equation 6.9 in equation 6.8, we get

`color{purple}(ΔH = ΔU + Δn_gRT)` ................ (6.10)

The equation 6.10 is useful for calculating `color{purple}(ΔH)` from `color{purple}(ΔU)` and vice versa.

● But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure. So, we need to define another state function which may be suitable under these conditions.

`=>` We may write equation (6.1) as `color{purple}(ΔU = q_p− pΔV)` at constant pressure, where `q_p` is heat absorbed by the system and `color{purple}(–pΔV)` represent expansion work done by the system.

● Let us represent the initial state by subscript `1` and final state by `2`. We can rewrite the above equation as

`color{purple}(U_2–U_1 = q_p – p (V_2 – V_1)`

● On rearranging, we get

`color{purple}(q_p = (U_2 + pV_2) – (U_1 + pV_1))` ........(6.6)

Now we can define another thermodynamic function, the enthalpy `H` [Greek word enthalpien, to warm or heat content] as :

`color{purple}(H = U + pV)` .................... (6.7)

So, equation (6.6) becomes

`color{purple}(q_p= H_2 – H_1 = ΔH)`

`=>` Although `color{purple}(q)` is a path dependent function, `color{purple}(H)` is a state function because it depends on `U`, `p` and `V`, all of which are state functions.

● Therefore, `color{purple}(ΔH)` is independent of path. Hence, `color{purple}(q_p)` is also independent of path.

`=>` For finite changes at constant pressure, we can write equation 6.7 as

`color{purple}(ΔH = ΔU + ΔpV)`

Since `color{purple}(p)` is constant, we can write

`color{purple}(ΔH = ΔU + pΔV)` .................... (6.8)

`color{red}("Note ")` When heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.

`=>` `color{purple}(ΔH = q_p)`, heat absorbed by the system at constant pressure.

● `color{purple}(ΔH)` is negative for `color{red}("exothermic reactions")` which evolve heat during the reaction

● `color{purple}(ΔH)` is positive for `color{red}("endothermic reactions")` which absorb heat from the surroundings.

`=>` At constant volume `color{purple}((ΔV = 0), ΔU = q_V),` therefore equation 6.8 becomes

`color{purple}(ΔH = ΔU = q_V)`

`=>` The difference between `color{purple}(ΔH)` and `color{purple}(ΔU)` is not usually significant for systems consisting of only solids and/or liquids. Solids and liquids do not suffer any significant volume changes upon heating.

`=>` The difference becomes significant when gases are involved.

● Let us consider a reaction involving gases.

● If `color{purple}(V_A)` is the total volume of the gaseous reactants, `color{purple}(V_B)` is the total volume of the gaseous products, `n_A` is the number of moles of gaseous reactants and `color{purple}(n_B)` is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,

`color{purple}(pV_A = n_ART)` and `color{purple}(pV_B = n_BRT)`

Thus, `color{purple}(pV_B – pV_A = n_BRT – n_ART = (n_B–n_A)RT)`

or `color{purple}(p (V_B – V_A) = (n_B – n_A) RT)`

or` color{purple}(p ΔV = Δn_gRT)` ................. (6.9)

● Here, `color{purple}(Δn_g)` refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.

● Substituting the value of `color{purple}(pΔV)` from equation 6.9 in equation 6.8, we get

`color{purple}(ΔH = ΔU + Δn_gRT)` ................ (6.10)

The equation 6.10 is useful for calculating `color{purple}(ΔH)` from `color{purple}(ΔU)` and vice versa.

Q 3047623583

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and `100°C` is `41kJ mol^(–1)`. Calculate the internal energy change, when

(i) 1 mol of water is vaporised at 1 bar pressure and `100°C.`

(ii) 1 mol of water is converted into ice.

(i) 1 mol of water is vaporised at 1 bar pressure and `100°C.`

(ii) 1 mol of water is converted into ice.

(i) The change `H_2O (l) → H_2O (g)`

`DeltaH = DeltaU+ Deltan_g RT`

or `DeltaU = DeltaH – Deltan_gR T` , substituting the values, we get

`DeltaU = 41.00kJ mol^(-1) -1 xx8.3 J mol^(-1) K^(-1) xx 373K`

` = 41.00kJ mol^(-1) -3.096 kJ mol^(-1) `

` = 37.904 kJ mol^(-1)`

(ii) The change `H_2O (l) → H_2O (s)`

There is negligible change in volume, So, we can put `p Delta V = Deltan_g RT approx O` in this case `DeltaH approx DeltaU`

so `DeltaU = 41.00kJ mol^(-1)`

`=>` In thermodynamics, a distinction is made between extensive properties and intensive properties.

`color{green}("Extensive Property ")` An extensive property is a property whose value depends on the quantity or size of matter present in the system.

`color{red}("Example ")` Mass, volume, internal energy, enthalpy, heat capacity etc. are extensive properties.

`color{green}("Intensive Properties ")` Those properties which do not depend on the quantity or size of matter present are known as intensive properties.

`color{red}("Example ")` Temperature, density, pressure etc. are intensive properties.

`=>` A molar property, `color{purple}(χ_m)`, is the value of an extensive property `color{purple}(χ)` of the system for `1` mol of the substance.

● If `n` is the amount of matter, `color{purple}(χ_m = χ/n)` is independent of the amount of matter.

● Other examples are molar volume, `color{purple}(V_m)` and molar heat capacity, `color{purple}(C_m)`.

`=>` The distinction between extensive and intensive properties can be understood by considering a gas enclosed in a container of volume` color{purple}(V)` and at temperature `color{purple}(T)` [Fig. 6.6(a)].

● Let us make a partition such that volume is halved, each part [Fig. 6.6 (b)] now has one half of the original volume, `V/2`, but the temperature will still remain the same i.e., `color{purple}(T)`.

● It is clear that volume is an extensive property and temperature is an intensive property.

`color{green}("Extensive Property ")` An extensive property is a property whose value depends on the quantity or size of matter present in the system.

`color{red}("Example ")` Mass, volume, internal energy, enthalpy, heat capacity etc. are extensive properties.

`color{green}("Intensive Properties ")` Those properties which do not depend on the quantity or size of matter present are known as intensive properties.

`color{red}("Example ")` Temperature, density, pressure etc. are intensive properties.

`=>` A molar property, `color{purple}(χ_m)`, is the value of an extensive property `color{purple}(χ)` of the system for `1` mol of the substance.

● If `n` is the amount of matter, `color{purple}(χ_m = χ/n)` is independent of the amount of matter.

● Other examples are molar volume, `color{purple}(V_m)` and molar heat capacity, `color{purple}(C_m)`.

`=>` The distinction between extensive and intensive properties can be understood by considering a gas enclosed in a container of volume` color{purple}(V)` and at temperature `color{purple}(T)` [Fig. 6.6(a)].

● Let us make a partition such that volume is halved, each part [Fig. 6.6 (b)] now has one half of the original volume, `V/2`, but the temperature will still remain the same i.e., `color{purple}(T)`.

● It is clear that volume is an extensive property and temperature is an intensive property.

`=>` In this section, let us see how to measure heat transferred to a system. This heat appears as a rise in temperature of the system in case of heat absorbed by the system.

● The increase of temperature is proportional to the heat transferred `color{purple}(q = text(coeff) × ΔT)`

● The magnitude of the coefficient depends on the size, composition and nature of the system.

● We can also write it as `color{purple}(q = C ΔT)`

● The coefficient, `color{purple}(C)` is called the `color{red}("heat capacity")`.

● Thus, we can measure the heat supplied by monitoring the temperature rise, provided we know the heat capacity.

● When `color{purple}(C)` is large, a given amount of heat results in only a small temperature rise. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature.

`color{purple}(C)` is directly proportional to amount of substance. The molar heat capacity of a substance, `C_m = C/n` is the heat capacity for one mole of the substance and is the quantity of heat needed to raise the temperature of one mole by one degree celsius (or one kelvin).

`color{green}("Specific Heat Capacity" )` Specific heat, also called specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree celsius (or one kelvin).

● For finding out the heat, `color{purple}(q)`, required to raise the temperatures of a sample, we multiply the specific heat of the substance, `color{purple}(c)`, by the mass `color{purple}(m)`, and temperatures change, `color{purple}(ΔT)` as

`color{purple}(q = c xx m xx Delta T)` .............(6.11)

● The increase of temperature is proportional to the heat transferred `color{purple}(q = text(coeff) × ΔT)`

● The magnitude of the coefficient depends on the size, composition and nature of the system.

● We can also write it as `color{purple}(q = C ΔT)`

● The coefficient, `color{purple}(C)` is called the `color{red}("heat capacity")`.

● Thus, we can measure the heat supplied by monitoring the temperature rise, provided we know the heat capacity.

● When `color{purple}(C)` is large, a given amount of heat results in only a small temperature rise. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature.

`color{purple}(C)` is directly proportional to amount of substance. The molar heat capacity of a substance, `C_m = C/n` is the heat capacity for one mole of the substance and is the quantity of heat needed to raise the temperature of one mole by one degree celsius (or one kelvin).

`color{green}("Specific Heat Capacity" )` Specific heat, also called specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree celsius (or one kelvin).

● For finding out the heat, `color{purple}(q)`, required to raise the temperatures of a sample, we multiply the specific heat of the substance, `color{purple}(c)`, by the mass `color{purple}(m)`, and temperatures change, `color{purple}(ΔT)` as

`color{purple}(q = c xx m xx Delta T)` .............(6.11)

`=>` At constant volume, the heat capacity, `color{purple}(C)` is denoted by `color{purple}(C_V)` and at constant pressure, this is denoted by `color{purple}(C_p)`.

● Let us find the relationship between the two.

`=>` We can write equation for heat, `color{purple}(q)`

● at constant volume as `color{purple}(q_V = C_VΔT = ΔU)`

● at constant pressure as `color{purple}(q_p = C_p DeltaT = Delta H)`

`=>` The difference between` color{purple}(C_p)` and `color{purple}(C_V)` can be derived for an ideal gas as :

● For a mole of an ideal gas, `color{purple}(ΔH = ΔU + Δ(pV ))`

`color{purple}(= ΔU + Δ(RT ))`

`color{purple}(= ΔU + RΔT)`

`∴ color{purple}(ΔH = ΔU + RΔT)` ................... (6.12)

On putting the values of `color{purple}(ΔH)` and `color{purple}(ΔU),` we have

`color{purple}(C_pΔT = C_VΔT+RΔT)`

`color{purple}(C_p = C_v + R)`

`color{purple}(C_p -C_v = R)` ......(6.13)

● Let us find the relationship between the two.

`=>` We can write equation for heat, `color{purple}(q)`

● at constant volume as `color{purple}(q_V = C_VΔT = ΔU)`

● at constant pressure as `color{purple}(q_p = C_p DeltaT = Delta H)`

`=>` The difference between` color{purple}(C_p)` and `color{purple}(C_V)` can be derived for an ideal gas as :

● For a mole of an ideal gas, `color{purple}(ΔH = ΔU + Δ(pV ))`

`color{purple}(= ΔU + Δ(RT ))`

`color{purple}(= ΔU + RΔT)`

`∴ color{purple}(ΔH = ΔU + RΔT)` ................... (6.12)

On putting the values of `color{purple}(ΔH)` and `color{purple}(ΔU),` we have

`color{purple}(C_pΔT = C_VΔT+RΔT)`

`color{purple}(C_p = C_v + R)`

`color{purple}(C_p -C_v = R)` ......(6.13)