### Related Formula Q = n times Delta H_textvap^0 quad textwhere n = fractextMass, textMolar Mass$Q = n \times \Delta H_{\text{vap}}^0 \quad \text{where } n = \frac{\text{Mass}}, {\text{Molar Mass}}$ ### Core Logic First, calculate the molar mass of carbon tetrachloride (CCl_4$CCl_4$): textMolar mass = 12 + 4(35.5) = 12 + 142 = 154text g/mol$\text{Molar mass} = 12 + 4(35.5) = 12 + 142 = 154\text{ g/mol}$ Next, calculate the total number of moles present in 284text g$284\text{ g}$ of the substance: n = frac284, 154 approx 1.844text moles$n = \frac{284}, {154} \approx 1.844\text{ moles}$ ### Step 1: Enthalpy Calculation Calculate the total energy required for vaporization: Delta H = 1.844text mol times 30.5text kJ/mol approx 56.24text kJ$\Delta H = 1.844\text{ mol} \times 30.5\text{ kJ/mol} \approx 56.24\text{ kJ}$ Rounding to the nearest integer value gives **56**. ### Pattern Recognition Enthalpy of vaporization is an intensive property given per mole. Scale it lineary by multiplying by the total number of moles to find the total extensive heat required. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Related Formula First law of thermodynamics framework: Delta U = Q + W$\Delta U = Q + W$ For an isothermal processes involving ideal gases, internal energy change is zero: Delta U = 0 implies Q = -W$\Delta U = 0 \implies Q = -W$ Irreversible work formula expanding against constant external pressure: W = -P_textext Delta V = -P_textext(V_2 - V_1)$W = -P_{\text{ext}} \Delta V = -P_{\text{ext}}(V_2 - V_1)$ ### Step 1: Calculate structural work values Given values: P_textext = 80text kPa = 80 times 10^3text Pa$P_{\text{ext}} = 80\text{ kPa} = 80 \times 10^3\text{ Pa}$ V_1 = 30text dm^3 = 30 times 10^-3text m^3$V_1 = 30\text{ dm}^3 = 30 \times 10^{-3}\text{ m}^3$ V_2 = 45text dm^3 = 45 times 10^-3text m^3$V_2 = 45\text{ dm}^3 = 45 \times 10^{-3}\text{ m}^3$ Delta V = (45 - 30) times 10^-3 = 15 times 10^-3text m^3$\Delta V = (45 - 30) \times 10^{-3} = 15 \times 10^{-3}\text{ m}^3$ W = -80 times 10^3 times (15 times 10^-3) = -1200text J$W = -80 \times 10^3 \times (15 \times 10^{-3}) = -1200\text{ J}$ ### Step 2: Solve for heat magnitude $Q = -W = -(-1200text J) = 1200text J$$Q = -W = -(-1200\text{ J}) = 1200\text{ J}$ ### Pattern Recognition Constant opposing pressure indicates an irreversible process path. Use $ ### Pattern Recognition Constant opposing pressure indicates an irreversible process path. Use $W = -P\Delta V$ instead of logarithmic integrals. ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Core Logic According to the second law of thermodynamics, at constant temperature and pressure, the change in Gibbs free energy (Delta G$\Delta G$) dictates the spontaneity of a process. - If Delta G lt 0$\Delta G \lt 0$ (negative), the process is spontaneous. - If Delta G gt 0$\Delta G \gt 0$ (positive), the process is non-spontaneous. - If Delta G = 0$\Delta G = 0$, the system is in equilibrium (reversible process). Therefore, the statement "Delta G$\Delta G$ is positive for a spontaneous reaction" is factually incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided. ### Core Logic Let the given reactions be: (1) 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), quad Delta H_1 = -822 \, mathrmkJ/mol$2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}, \quad \Delta H_1 = -822 \, \mathrm{kJ/mol}$ (2) mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), quad Delta H_2 = -110 \, mathrmkJ/mol$\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}, \quad \Delta H_2 = -110 \, \mathrm{kJ/mol}$ Target Reaction (3): 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g), quad Delta H_3 = ?$3\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow 2\mathrm{Fe}_{(s)} + 3\mathrm{CO}_{(g)}, \quad \Delta H_3 = ?$ To construct the target reaction: - We need 3 mathrmCO_(g)$3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3. - We need mathrmFe_2mathrmO_3(s)$\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and 2 mathrmFe_(s)$2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1). ### Step 1: Calculate Net Enthalpy Target Reaction (3) = 3 times (2) - (1)$3 \times (2) - (1)$ Delta H_3 = 3 times Delta H_2 - Delta H_1$\Delta H_3 = 3 \times \Delta H_2 - \Delta H_1$ Delta H_3 = 3(-110) - (-822)$\Delta H_3 = 3(-110) - (-822)$ Delta H_3 = -330 + 822 = 492 \, mathrmkJ/mol$\Delta H_3 = -330 + 822 = 492 \, \mathrm{kJ/mol}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula W_textcyclic = textArea enclosed in P-V graph$W_{\text{cyclic}} = \text{Area enclosed in P-V graph}$ ### Core Logic The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume (V$V$) on the y-axis versus Pressure (P$P$) on the x-axis. The path A rightarrow B rightarrow C rightarrow A$A \rightarrow B \rightarrow C \rightarrow A$ is traced in a clockwise direction on the V$V$-P$P$ graph. Clockwise on a V$V$-P$P$ graph corresponds to anti-clockwise on a standard P$P$-V$V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for). ### Step 1: Calculating Area The enclosed region is a right-angled triangle. Base of triangle on P-axis = 30 - 10 = 20 text kPa$30 - 10 = 20 \text{ kPa}$ Height of triangle on V-axis = 30 - 10 = 20 text dm^3$30 - 10 = 20 \text{ dm}^3$ textArea = frac12 times textbase times textheight$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ textArea = frac12 times 20 times 20 = 200 text kPacdottextdm^3$\text{Area} = \frac{1}{2} \times 20 \times 20 = 200 \text{ kPa}\cdot\text{dm}^3$ ### Step 2: Unit conversion 1 text kPa = 10^3 text Pa$1 \text{ kPa} = 10^3 \text{ Pa}$ 1 text dm^3 = 1 text Litre = 10^-3 text m^3$1 \text{ dm}^3 = 1 \text{ Litre} = 10^{-3} \text{ m}^3$ W = 200 times 10^3 text Pa times 10^-3 text m^3$W = 200 \times 10^3 \text{ Pa} \times 10^{-3} \text{ m}^3$ W = 200 text J$W = 200 \text{ J}$ ### Pattern Recognition 1 text kPa cdot 1 text L = 1 text Joule$1 \text{ kPa} \cdot 1 \text{ L} = 1 \text{ Joule}$. This direct conversion saves time without converting explicitly to standard SI units (Pa$Pa$ and m^3$m^3$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics