UNIT-I
1. (a) Define state and path functions, and give one example of each. 2
State Function: A thermodynamic property whose value depends only on the current state of the system (initial and final states) and is independent of the path taken to reach that state
.Path Function: A thermodynamic property whose value depends on the specific path or process followed during the change between states
.1. (b) State and explain the zeroth law of thermodynamics. 2
The Zeroth Law of Thermodynamics states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
Explanation: If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are at the same temperature. This law provides the basis for the measurement of temperature using thermometers.
1. (c) What are exact and inexact differentials? 2
Exact Differential: A differential of a state function (like dU or dH) where the integral depends only on the initial and final points. Mathematically, it satisfies Euler's reciprocity relation.
Inexact Differential: A differential of a path function (like dw or dq) where the integral depends on the path taken. These are often denoted by a cross or delta symbol (δw, δq).
2. (a) (i) Distinguish between isothermal and adiabatic process. 2
2. (a) (ii) Derive a relation between temperature and volume in reversible adiabatic expansion. 3
For an adiabatic process, dq = 0. From First Law: dU = dw = -P dV
.For 1 mole of an ideal gas, dU = Cv dT and P = RT/V
.Cv dT = -(RT/V) dV
(Cv/R) (dT/T) = -(dV/V)
Integrating both sides between (T1, V1) and (T2, V2):
(Cv/R) ln(T2/T1) = -ln(V2/V1) = ln(V1/V2)
Since R = Cp - Cv, then R/Cv = (Cp/Cv) - 1 = γ - 1
.T V^(γ-1) = constant
2. (a) (iii) One mole of an ideal gas expands against a constant external pressure of 1 atm from 10 dm³ to 30 dm³. Calculate work. 2
Work done in irreversible expansion against constant external pressure (Pext):
w = -Pext (V2 - V1)
w = -1 atm * (30 dm³ - 10 dm³) = -1 atm * 20 dm³
To convert to Joules (1 L-atm = 101.325 J):
w = -20 * 101.325 J = -2026.5 J
2. (a) (iv) Derive the expression for the work done in reversible isothermal expansion. 3
In a reversible isothermal process, T is constant and P = nRT/V
.w = -∫ P dV (from V1 to V2)
w = -∫ (nRT/V) dV = -nRT ∫ (1/V) dV
w = -nRT [ln V] (from V1 to V2)
w = -nRT ln(V2/V1) or w = -2.303 nRT log(V2/V1)
2. (b) (i) Calculate enthalpy of formation of glucose. 2
Combustion of Glucose: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
ΔH_comb = [6 * ΔHf(CO2) + 6 * ΔHf(H2O)] - [ΔHf(C6H12O6) + 6 * ΔHf(O2)]
-2816 = [6 * (-393.5) + 6 * (-285.9)] - [ΔHf(C6H12O6) + 0]
-2816 = [-2361 - 1715.4] - ΔHf(C6H12O6)
-2816 = -4076.4 - ΔHf(C6H12O6)
ΔHf(C6H12O6) = -4076.4 + 2816
ΔHf(C6H12O6) = -1260.4 kJ/mol
2. (b) (ii) Deduce Kirchhoff's equation. 3
Kirchhoff's equation describes the variation of enthalpy of reaction with temperature
.ΔH = H(products) - H(reactants)
Differentiating with respect to T at constant pressure:
(d(ΔH)/dT)p = (dH(prod)/dT)p - (dH(react)/dT)p
Since (dH/dT)p = Cp:
(d(ΔH)/dT)p = ΔCp
Integrating between T1 and T2:
ΔH2 - ΔH1 = ΔCp (T2 - T1)
UNIT-II
3. (a) Give the statement of second law of thermodynamics in terms of entropy. 2
The entropy of the universe (an isolated system) increases in the course of any spontaneous change.
Mathematically, for a spontaneous process: ΔS_total > 0
.4. (a) (ii) Derive Gibbs-Helmholtz equation for a process at constant pressure. 3
We know G = H - TS. For a process: ΔG = ΔH - TΔS.
From Maxwell relations, (dΔG/dT)p = -ΔS
.Substituting -ΔS in the ΔG equation:
ΔG = ΔH + T(dΔG/dT)p
Rearranging this gives the Gibbs-Helmholtz equation:
[d(ΔG/T) / dT]p = -ΔH / T²
4. (b) (ii) Show that -ΔA_T = w_max. 2
Work function A is defined as A = U - TS. At constant temperature: ΔA = ΔU - TΔS.
From Second Law, for a reversible process: ΔS = q_rev / T, so TΔS = q_rev
.ΔA = ΔU - q_rev. From First Law: ΔU - q = w.
Therefore, ΔA = w_rev
.Since work done by the system is -w, then -ΔA = -w_rev = w_max.
The decrease in work function (Helmholtz free energy) at constant temperature equals the maximum work obtainable from the system.
UNIT-III
5. (a) What is triple point? 2
The triple point is the specific temperature and pressure at which all three phases (solid, liquid, and gas) of a substance coexist in thermodynamic equilibrium
. For water, the triple point is at 273.16 K and 0.006 atm.6. (b) (iii) Explain the phase diagram of CO2 in detail. 3
The phase diagram of CO2 consists of three areas representing Solid, Liquid, and Gas phases
.UNIT-IV
7. (a) State and explain Le Chatelier's principle. 2
If a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will shift its equilibrium position in a direction that tends to counteract the effect of the change.
Example: In the Haber process (exothermic), increasing pressure shifts equilibrium toward ammonia production to reduce the number of moles
.8. (a) (i) Deduce Gibbs-Duhem equation. 2
The total Gibbs free energy of a mixture is G = Σ ni μi
.Differentiating G: dG = Σ ni dμi + Σ μi dni
.Also, from the fundamental equation at constant T and P: dG = Σ μi dni
.Comparing the two equations:
Σ ni dμi = 0
This shows that chemical potentials of components in a mixture are not independent
.UNIT-V
9. (a) Define pH. Calculate pH of 10⁻⁹ M HCl. 4
pH Definition: pH is the negative logarithm (base 10) of the molar concentration of hydrogen ions: pH = -log[H+]
.Calculation: Since the concentration is very dilute (10⁻⁹ M), we must include [H+] from water (10⁻⁷ M)
.[H+]_total = [H+]_acid + [H+]_water ≈ 1.01 * 10⁻⁷ M.
pH = -log(1.01 * 10⁻⁷) ≈ 6.99
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