Given: Δ Hsub[C(graphite)] = 710 kJ mol⁻¹$$\Delta H_{\text{sub}}^{\ominus}[\text{C}(\text{graphite})] = 710\text{ kJ mol}^{-1}$$ ΔC-HH = 414 kJ mol⁻¹$$\Delta_{\text{C}-\text{H}}H^{\ominus} = 414\text{ kJ mol}^{-1}$$ ΔH-HH = 436 kJ mol⁻¹$$\Delta_{\text{H}-\text{H}}H^{\ominus} = 436\text{ kJ mol}^{-1}$$ ΔC=CH = 611 kJ mol⁻¹$$\Delta_{\text{C}=\text{C}}H^{\ominus} = 611\text{ kJ mol}^{-1}$$ The Δ Hf$\Delta H_{\text{f}}^{\ominus}$ for CH₂=CH₂$\text{CH}_2=\text{CH}_2$ is ________ kJ mol⁻¹$\text{kJ mol}^{-1}$ (nearest integer value)

Formulas Used

For an ideal gas undergoing a constant pressure process (1.00 atm$1.00\text{ atm}$):

Change in internal energy (Δ U$\Delta U$):

For a monoatomic gas like Argon:

• Cv = (3)/(2) R$C_v = \frac{3}{2} R$
• Cₚ = (5)/(2) R$C_p = \frac{5}{2} R$

Core Logic

Step 1: Calculate the final temperature (Tf$T_f$) Heat transferred at constant pressure (qₚ$q_p$) = 500 J$500\text{ J}$

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Step 2: Calculate the change in internal energy (Δ U$\Delta U$)

Alternatively, using the ratio of heat capacities:

Thus, the final temperature is 348 K$348\text{ K}$ and the change in internal energy is 300 J$300\text{ J}$.

Pattern Recognition

For a monoatomic ideal gas under constant pressure, exactly 60%$60\%$ of the heat added ((Cv)/(Cₚ) = (3)/(5)$\frac{C_v}{C_p} = \frac{3}{5}$) goes into increasing the internal energy (Δ 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 (Pₑₓₜ = 0$P_{ext} = 0$). Work done: w = -Pₑₓₜ Δ V$w = -P_{ext} \Delta V$. Since Pₑₓₜ = 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, Δ U = q + w$\Delta U = q + w$. Since q = 0$q = 0$ and w = 0$w = 0$, the change in internal energy Δ U = 0$\Delta U = 0$.

For an ideal gas, internal energy is a function of temperature only (Δ U = nCvΔ T$\Delta U = nC_v\Delta T$). If Δ U = 0$\Delta U = 0$, then Δ T = 0$\Delta T = 0$.

Step 1: Final Parameter Check

Evaluating all parameters simultaneously: q = 0$q = 0$ w = 0$w = 0$ Δ T = 0$\Delta T = 0$

Pattern Recognition

Adiabatic + Free Expansion of IDEAL gas arrow$\rightarrow$ Nothing changes thermodynamically except volume and pressure. q = 0, w = 0, Δ U = 0, Δ T = 0, Δ H = 0$q = 0, w = 0, \Delta U = 0, \Delta T = 0, \Delta H = 0$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Related Formula

Core Logic

First, calculate the molar mass of carbon tetrachloride (CCl₄$\mathrm{CCl}_4$):

Next, calculate the total number of moles present in 284 g$284\text{ g}$ of the substance:

Step 1: Enthalpy Calculation

Calculate the total energy required for vaporization:

Rounding to the nearest integer value gives 56.

Pattern Recognition

Enthalpy of vaporization is an intensive property given per mole. Scale it linearly by multiplying by the total number of moles to find the total extensive heat required.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Related Formula

First law of thermodynamics framework:

For an isothermal processes involving ideal gases, internal energy change is zero:

Irreversible work formula expanding against constant external pressure:

Step 1: Calculate structural work values

Given values:

Step 2: Solve for heat magnitude

Core Logic

According to the second law of thermodynamics, at constant temperature and pressure, the change in Gibbs free energy (Δ G$\Delta G$) dictates the spontaneity of a process.

• If Δ G lt 0$\Delta G \lt 0$ (negative), the process is spontaneous.
• If Δ G gt 0$\Delta G \gt 0$ (positive), the process is non-spontaneous.
• If Δ G = 0$\Delta G = 0$, the system is in equilibrium (reversible process).

Therefore, the statement "Δ G$\Delta G$ is positive for a spontaneous reaction" is factually incorrect.

Chapter Mix

Class 11 Chemistry: Thermodynamics