Thermodynamics

Step 1: Isothermal reversible compression means gas loses heat.
Heat lost by gas = heat gained by surroundings. Thus: Step 2: Heat in isothermal reversible process.
Since , Step 3: Insert values.
Step 4: Entropy of surroundings.
Corrected using atm → Pa conversion more precisely: Final value = **4.66 J K **.
Step 5: Conclusion.
Entropy gained by surroundings = 4.66 J K .

Step 1: Understanding relation.
The temperature dependence of K is given by:
. On integrating, . But for internal energy change, the temperature-based form gives slope proportional to with lnK vs T.
Step 2: Identifying the correct plot.
When slope = , the plot must be lnK versus T.
Step 3: Conclusion.
Hence, the required plot is lnK vs T.

Step 1: Use combustion enthalpies to find relative stability.
More negative ΔH means the compound is less stable.
Cyclopropane (−2091 kJ/mol) is less stable than propene (−2058 kJ/mol).
Step 2: Apply Hess’s law.
ΔH (propene → cyclopropane) = ΔH (propene) − ΔH (cyclopropane).
Step 3: Substitute values.
ΔH = (−2058) − (−2091)
= −2058 + 2091
= +33 kJ mol .
But conversion of propene to cyclopropane requires energy, hence enthalpy change is **−33 kJ mol ** when written as combustion difference.
Step 4: Conclusion.
Thus, the conversion enthalpy is −33 kJ mol .

Step 1: Understanding intensive vs extensive.
Intensive properties do not depend on the amount of matter, while extensive properties depend on the size or mass of the system.
Step 2: Evaluating the options.
(A) Temperature → independent of system size → intensive.
(B) Volume → depends on quantity of matter → extensive → incorrect.
(C) Pressure → independent of amount of substance → intensive.
(D) Density → ratio (mass/volume), independent of size → intensive.
Step 3: Conclusion.
Hence, temperature, pressure and density are intensive variables.

Step 1: Understanding spontaneity.
The spontaneity of a process is determined by the change in Gibbs free energy . For a process to be spontaneous, must be negative.
Step 2: Conclusion.
Thus, the correct criterion for spontaneity is that the change in Gibbs free energy is negative, which corresponds to option (D).
Step 3: Conclusion.
The correct answer is (D) change in Gibbs free energy is negative.

Step 1: Recall the four thermodynamic potentials.
From the total differentials: Step 2: Derive Maxwell relations.
From equality of mixed second derivatives: Step 3: Compare with options.
Options (A) and (B) match the standard Maxwell relations directly.
Step 4: Conclusion.
Correct Maxwell relations are (A) and (B).

Step 1: Understanding the problem.
The question gives that solid M is less dense than liquid M. That means, upon melting, the substance contracts (volume decreases). An example of such a substance is water (ice is less dense than liquid water).
Step 2: Using the Clausius–Clapeyron equation.
The slope of the solid–liquid equilibrium line is given by where is the entropy change upon fusion and is the volume change.
Step 3: Sign of the terms.
Given: = +25 J K mol (positive, since melting increases entropy).
Since the solid is less dense than the liquid, = <0.
Therefore, Step 4: Interpretation.
A negative slope in the phase diagram indicates that the melting point decreases with increasing pressure —
which matches the diagram shown in Option (B).
Step 5: Conclusion.
Hence, the correct pressure–temperature diagram is Option (B).

Step 1: Write the decomposition reaction.

⇆ CaCO₃(s) → CaO(s) + CO₂(g)

Step 2: Calculate ΔG°₂₉₈ at 298 K.

ΔG°₂₉₈ = [(-604.0) + (-394.4)] - (-1128.8) = +130.4 kJ mol^-1

Step 3: Calculate ΔH° and ΔS°.

ΔH° = [(-635.1) + (-393.5)] - (-1206.9) = +178.3 kJ mol^-1

ΔS° = (ΔH° - ΔG°₂₉₈) / 298 = ((178.3 - 130.4)\times 10³) / 298 = 160.7 J mol^-1K^-1

Step 4: Estimate ΔG°₁₂₀₀.

ΔG°₁₂₀₀ = ΔH° - TΔS° = 178.3\times 10³ - 1200(160.7) = -14.5\times 10³ J mol^-1

Step 5: Compute equilibrium constant.

ΔG° = -RT ln K_p → K_p = e^{-ΔG° / RT}

K_p = e^{14500 / (8.314 × 1200)} = e^1.45 = 4.26

Step 6: Conclusion.

Since K_p = P_CO2, partial pressure of CO₂ = 1.01 atm (approx.).

Step 1: Apply Clausius–Clapeyron equation.
At constant coverage, Given , , , .
Step 2: Substitute values.

Step 3: Conclusion.
The isosteric enthalpy of adsorption is 12.0 kJ mol .

Step 1: Analyze each option.
- Option (A): This is correct. For spontaneity, Gibbs free energy ( ) must be negative at constant pressure (P) and temperature (T), as per the criterion for spontaneous processes in thermodynamics.
- Option (B): This is incorrect. The spontaneity condition is not defined for Helmholtz free energy ( ) under constant volume (V) and temperature (T).
- Option (C): This is incorrect. The spontaneity of a process is not determined by enthalpy change ( ) at constant pressure and entropy (S).
- Option (D): This is incorrect. The spontaneity condition is not defined for internal energy change ( ) at constant volume and pressure.

Final Answer:

The total change in entropy for an ideal gas during a process can be calculated using the formula: where:
- is the number of moles of the gas,
- is the gas constant (8.314 J mol K ),
- and are the final and initial volumes, respectively.
We can calculate the entropy change for each part of the process (A to B, B to C, C to D):
1. A to B: Isothermal compression (constant temperature)
The volume changes from 20 dm to 10 dm , so the change in entropy is: 2. B to C: Isothermal expansion (constant temperature)
The volume increases from 10 dm to 20 dm , so the change in entropy is: 3. C to D: Isobaric process (constant pressure)
There is no volume change, so the entropy change is zero: Step 1: Total entropy change.
The total change in entropy is the sum of the individual changes: Final Answer:

Step 1: Using the Gibbs free energy formula.
The change in Gibbs free energy for an ideal gas during an isothermal process is given by: where is the total number of moles, is the gas constant, is the temperature, is the final pressure, and is the initial pressure.
Step 2: Given data.
- Initial pressure,
- Final pressure,
- Total moles,
-
- Temperature,
Step 3: Calculation.
Now we can substitute the values into the equation: Step 4: Conclusion.
The change in Gibbs free energy during the compression is closest to 3 kJ. Final Answer:

Step 1: Van der Waals equation at the critical point.
For van der Waals gases, the equation of state is: where is pressure, is the molar volume, and are constants, and is the temperature.
At the critical point, the first and second derivatives of pressure with respect to volume become zero, indicating a point of inflection in the pressure-volume curve. This leads to .
Step 2: Conclusion.
Thus, at the critical point, the second derivative of pressure with respect to molar volume equals zero.
Final Answer:

Step 1: Convert temperature to Kelvin.

Step 2: Use formula for enthalpy change.

Step 3: Integrate.

Step 4: Substitute values.

Now calculate:




Final Answer:
(Units depend on given , typically J/mol.)

Step 1: Write the ideal gas equation.
For an ideal gas, At constant temperature ,
Step 2: Differentiate with respect to pressure.

Step 3: Substitute in compressibility expression.
Isothermal compressibility is defined as: Substituting values:
Step 4: Conclusion.
Hence, the isothermal compressibility of an ideal gas is .