Thermodynamics Function Class 11 NCERT Notes & Practice

Understanding State Functions vs Path Functions

In thermodynamics, physical properties of a system are divided into two main categories: State Functions and Path Functions.

A State Function is a property whose value depends only on the current state of the system (characterized by pressure, temperature, volume, etc.) and is completely independent of the path or method used to reach that state. Examples of state functions include Temperature ($T$), Pressure ($P$), Volume ($V$), Internal Energy ($U$), Enthalpy ($H$), Entropy ($S$), and Gibbs Free Energy ($G$). If you change a system from state A to state B, the change in a state function ($\Delta X = X_B - X_A$) remains identical regardless of whether you carried out the process in a single step or over multiple intermediate steps.

Conversely, a Path Function is a thermodynamic property whose value depends on the specific path taken during a transition between states. Heat ($q$) and Work ($w$) are classic examples. For instance, the expansion of a gas from volume $V_1$ to $V_2$ can yield different amounts of work depending on whether the process is reversible or irreversible. Thus, while $\Delta U = q + w$ is a state function change, $q$ and $w$ individually are not.

Key Thermodynamic Functions Defined

Internal Energy (U)

The total energy stored within a thermodynamic system, comprising kinetic energy (due to molecular motions) and potential energy (due to chemical bonds and intermolecular forces).

Enthalpy (H)

The total heat content of a system at constant pressure, mathematically defined as $H = U + PV$.

Entropy (S)

A thermodynamic state function that serves as a measure of the molecular disorder or randomness within a system.

Gibbs Free Energy (G)

The thermodynamic potential used to calculate the maximum reversible work that may be performed by a system at constant temperature and pressure, defined as $G = H - TS$.

Deriving the Relationship Between Delta H and Delta U

1. Define Enthalpy — Start with the absolute definition of Enthalpy: $H = U + PV$.
2. Formulate the Change — At constant pressure, a change in Enthalpy ($\Delta H$) is written as: $\Delta H = \Delta U + P\Delta V$.
3. Apply Ideal Gas Law — For gaseous reactions, ideal gas behavior dictates $PV = nRT$. At constant temperature and pressure, $P\Delta V = \Delta n_g RT$, where $\Delta n_g$ is the difference in stoichiometric moles of gaseous products and gaseous reactants.
4. Substitute and Finalize — Substitute $P\Delta V$ back into the enthalpy equation to yield the vital CBSE formula: $\Delta H = \Delta U + \Delta n_g RT$.

Step-by-Step Solved Examples

• Example 1: Calculate $\Delta H$ for the reaction: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ at $298 \text{ K}$ given that $\Delta U = -92.2 \text{ kJ}$ and $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$. Step 1: Identify gaseous moles. $\Delta n_g = \text{Moles of gaseous products} - \text{moles of gaseous reactants} = 2 - (1 + 3) = -2$. Step 2: Use the formula $\Delta H = \Delta U + \Delta n_g RT$. Step 3: Convert units to ensure compatibility. $\Delta U = -92.2 \text{ kJ} = -92200 \text{ J}$. Step 4: Calculate the correction term: $\Delta n_g RT = (-2) \times 8.314 \times 298 = -4955.1 \text{ J} = -4.96 \text{ kJ}$. Step 5: Compute final enthalpy: $\Delta H = -92.2 \text{ kJ} + (-4.96 \text{ kJ}) = -97.16 \text{ kJ}$.
• Example 2: A reaction has $\Delta H = -110 \text{ kJ}$ and $\Delta S = -150 \text{ J K}^{-1}$. Determine if this reaction is spontaneous at $300 \text{ K}$. Step 1: Write the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$. Step 2: Standardize units. Convert $\Delta S$ to kJ: $\Delta S = -150 / 1000 = -0.150 \text{ kJ K}^{-1}$. Step 3: Substitute values: $\Delta G = -110 - (300 \times -0.150) = -110 + 45 = -65 \text{ kJ}$. Step 4: Analyze sign of $\Delta G$. Since $\Delta G$ is negative ($-65 \text{ kJ}$), the reaction is spontaneous at $300 \text{ K}$.

Important CBSE Board Sign Conventions

1. Work Sign: Under IUPAC convention, work done on the system (compression) is positive ($+w$), and work done by the system (expansion) is negative ($-w$).
2. Heat Sign: Heat absorbed by the system is positive ($+q$), and heat released by the system is negative ($-q$).
3. Unit Consistency: The gas constant $R$ is typically given as $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$. Ensure you convert your Enthalpy ($\Delta H$) and Internal Energy ($\Delta U$) from kilojoules (kJ) to joules (J), or divide the $RT$ term by 1000 before adding them together! Ignoring this step is the most common way students lose marks.

Practice Questions with Solutions

• Q: Under what thermodynamic condition is the change in enthalpy equal to the change in internal energy ($\Delta H = \Delta U$)? A: Step 1: Examine the relation $\Delta H = \Delta U + P\Delta V$ or $\Delta H = \Delta U + \Delta n_g RT$. Step 2: Identify conditions where the boundary term becomes zero. Step 3: This occurs if: (i) The reaction takes place in a closed vessel at constant volume (so $\Delta V = 0$), or (ii) There are no gaseous reactants or products involved in the reaction, or (iii) The change in the number of gaseous moles is zero ($\Delta n_g = 0$). Final answer: $\Delta H = \Delta U$ when $\Delta V = 0$ or $\Delta n_g = 0$.
• Q: State whether Volume ($V$) and Work ($w$) are state functions or path functions. A: Step 1: Recall the definition of a state function. It depends only on initial and final coordinates. Step 2: Volume depends solely on the current state configuration, so it is a State Function. Step 3: Recall the definition of a path function. It depends on the transition path. Step 4: Work done changes depending on whether expansion occurs against a constant external pressure or reversibly, so Work is a Path Function. Final answer: Volume is a state function; Work is a path function.
• Q: Calculate the value of $\Delta n_g$ for the combustion of liquid benzene: $C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$. A: Step 1: Identify all substances in the gaseous phase. $O_2(g)$ and $CO_2(g)$ are gases. Benzene and water are liquids, so they are excluded from $\Delta n_g$. Step 2: Calculate moles of gaseous products: $n_p = 6$. Step 3: Calculate moles of gaseous reactants: $n_r = \frac{15}{2} = 7.5$. Step 4: Find difference: $\Delta n_g = 6 - 7.5 = -1.5$. Final answer: $\Delta n_g = -1.5$.
• Q: Predict the spontaneity of a process where both $\Delta H$ and $\Delta S$ are positive. A: Step 1: Use the equation $\Delta G = \Delta H - T\Delta S$. Step 2: Since both terms are positive, $\Delta G = (+\Delta H) - T(+\Delta S)$. Step 3: For the process to be spontaneous, $\Delta G$ must be negative. Step 4: This is only possible if the magnitude of $T\Delta S$ is greater than $\Delta H$, which occurs at high temperatures. Final answer: The process is spontaneous only at high temperatures.

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