A sample of n-octane (1.14mathrm~g$1.14\mathrm{~g}$) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5mathrm~kJ~K^-1$5\mathrm{~kJ~K^{-1}}$. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is ________ mathrmkJ~mol^-1$\mathrm{kJ~mol^{-1}}$. (nearest integer)

### Related Formula Enthalpy of reaction from enthalpy of formation data: Delta H_textreaction = sum Delta H_f(textProducts) - sum Delta H_f(textReactants)$$\Delta H_{\text{reaction}} = \sum \Delta H_f(\text{Products}) - \sum \Delta H_f(\text{Reactants})$$ ### Core Logic Write out the balanced thermochemical equation for the combustion of benzene (C_6H_6$C_6H_6$): C_6H_6(l) + frac152O_2(g) rightarrow 6CO_2(g) + 3H_2O(l)$$C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$$ Given parameters: - Delta H_c = -3267.0mathrm\ kJ/mol$\Delta H_c = -3267.0\mathrm{\ kJ/mol}$ - Delta H_f[CO_2] = -393.5mathrm\ kJ/mol$\Delta H_f[CO_2] = -393.5\mathrm{\ kJ/mol}$ - Delta H_f[H_2O] = -286.0mathrm\ kJ/mol$\Delta H_f[H_2O] = -286.0\mathrm{\ kJ/mol}$ - Delta H_f[O_2] = 0mathrm\ kJ/mol$\Delta H_f[O_2] = 0\mathrm{\ kJ/mol}$ ### Step 1: Applying Hess's Law Substitute these values into the reaction expression: -3267 = [6(-393.5) + 3(-286.0)] - Delta H_f[C_6H_6]$$-3267 = [6(-393.5) + 3(-286.0)] - \Delta H_f[C_6H_6]$$ -3267 = [-2361.0 - 858.0] - Delta H_f[C_6H_6]$$-3267 = [-2361.0 - 858.0] - \Delta H_f[C_6H_6]$$ -3267 = -3219.0 - Delta H_f[C_6H_6]$$-3267 = -3219.0 - \Delta H_f[C_6H_6]$$ Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48mathrm\ kJ/mol$$\Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48\mathrm{\ kJ/mol}$$ ### Pattern Recognition Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply CO_2$CO_2$ by 6$6$ and H_2O$H_2O$ by 3$3$) to ensure accurate bookkeeping. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Formulas Used For an ideal gas undergoing a constant pressure process (1.00text atm$1.00\text{ atm}$): q_p = n cdot C_p cdot Delta T$$q_p = n \cdot C_p \cdot \Delta T$$ Change in internal energy (Delta U$\Delta U$): Delta U = n cdot C_v cdot Delta T$$\Delta U = n \cdot C_v \cdot \Delta T$$ For a monoatomic gas like Argon: * C_v = frac32 R$C_v = \frac{3}{2} R$ * C_p = frac52 R$C_p = \frac{5}{2} R$ ### Core Logic **Step 1: Calculate the final temperature (T_f$T_f$)** Heat transferred at constant pressure (q_p$q_p$) = 500text J$500\text{ J}$ 500 = 0.5 times left(frac52 times 8.3right) times (T_f - 298)$$500 = 0.5 \times \left(\frac{5}{2} \times 8.3\right) \times (T_f - 298)$$ 500 = 0.5 times 20.75 times (T_f - 298)$$500 = 0.5 \times 20.75 \times (T_f - 298)$$ 500 = 10.375 times (T_f - 298)$$500 = 10.375 \times (T_f - 298)$$ T_f - 298 = frac50010.375 approx 48.2text K$$T_f - 298 = \frac{500}{10.375} \approx 48.2\text{ K}$$ T_f = 298 + 48.2 = 346.2text K approx 348text K$$T_f = 298 + 48.2 = 346.2\text{ K} \approx 348\text{ K}$$ --- **Step 2: Calculate the change in internal energy (Delta U$\Delta U$)** Delta U = n cdot C_v cdot Delta T$$\Delta U = n \cdot C_v \cdot \Delta T$$ Alternatively, using the ratio of heat capacities: Delta U = left(fracC_vC_pright) times q_p = frac35 times 500text J = 300text J$$\Delta U = \left(\frac{C_v}{C_p}\right) \times q_p = \frac{3}{5} \times 500\text{ J} = 300\text{ J}$$ Thus, the final temperature is **348text K$348\text{ K}$** and the change in internal energy is **300text J$300\text{ J}$**. ### Pattern Recognition For a monoatomic ideal gas under constant pressure, exactly 60\%$60\%$ of the heat added (fracC_vC_p = frac35$\frac{C_v}{C_p} = \frac{3}{5}$) goes into increasing the internal energy (Delta U$\Delta U$), while 40\%$40\%$ is lost to expansion work (W$W$). **Correct Option:** **(A)**

### Core Logic Free expansion means expansion against a vacuum (P_ext = 0$P_{ext} = 0$). Work done: w = -P_ext Delta V$w = -P_{ext} \Delta V$. Since P_ext = 0$P_{ext} = 0$, w = 0$w = 0$. Adiabatic condition means there is no heat exchange with the surroundings. Heat transfer: q = 0$q = 0$. According to the First Law of Thermodynamics, Delta U = q + w$\Delta U = q + w$. Since q = 0$q = 0$ and w = 0$w = 0$, the change in internal energy Delta U = 0$\Delta U = 0$. For an ideal gas, internal energy is a function of temperature only (Delta U = nC_vDelta T$\Delta U = nC_v\Delta T$). If Delta U = 0$\Delta U = 0$, then Delta T = 0$\Delta T = 0$. ### Step 1: Final Parameter Check Evaluating all parameters simultaneously: q = 0$q = 0$ w = 0$w = 0$ Delta T = 0$\Delta T = 0$ ### Pattern Recognition Adiabatic + Free Expansion of IDEAL gas rightarrow$\rightarrow$ Nothing changes thermodynamically except volume and pressure. q = 0, w = 0, Delta U = 0, Delta T = 0, Delta H = 0$q = 0, w = 0, \Delta U = 0, \Delta T = 0, \Delta H = 0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### 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