Which of the following properties will change when system containing solution 1 will become solution 2?
Solution state transition grid for Q35 - JEE Main 2025 Morning
The flowchart maps Solution 1 containing 10 mol solute in 10 L water transitioning to Solution 2 containing 1 mol solute in 1 L water.

Solution & Explanation

Core Logic

Let us compute the concentration of both solutions:

Concentration of Solution 1 = 10 mol10 L = 1 mol/L Concentration of Solution 2 = 1 mol1 L = 1 mol/L

Since concentration is identical, both systems share matching compositions. Consequently, all intensive properties (independent of mass/size) like concentration, density, and molar heat capacity remain exactly equal.

Step 1: Identifying the Variable

Gibbs free energy (G) is an extensive property that scales directly with the amount of matter in the system. Because Solution 1 contains a larger total mass and volume than Solution 2, its overall Gibbs free energy value will change.

Pattern Recognition

Shortcut: Look for the only extensive property in the options. Density, concentration, and molar parameters are always intensive. Gibbs free energy (G) scales with total matter quantity.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

More Chemical Thermodynamics Previous-Year Questions — Page 9

Q83 jee_main_2024_30_january_evening Hess's Law of Constant Heat Summation
Two reactions are given below: 2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H° = -822 kJ/mol C(s) + (1)/(2)O2(g) arrow CO(g), Δ H° = -110 kJ/mol Then enthalpy change for following reaction 3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g)
Numerical Answer. Answer: 492 to 492

Solution

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) 2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H₁ = -822 kJ/mol (2) C(s) + (1)/(2)O2(g) arrow CO(g), Δ H₂ = -110 kJ/mol

Target Reaction (3):

3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g), Δ H₃ = ?

To construct the target reaction:

  • We need 3 CO(g) on the product side, so we multiply reaction (2) by 3.
  • We need Fe₂O3(s) on the reactant side and 2 Fe(s) on the product side, so we reverse reaction (1).
Step 1: Calculate Net Enthalpy

Target Reaction (3) = 3 × (2) - (1)

Δ H₃ = 3 × Δ H₂ - Δ H₁ Δ H₃ = 3(-110) - (-822) Δ H₃ = -330 + 822 = 492 kJ/mol
Chapter Mix

Class 11 Chemistry: Thermodynamics

Q jee_main_2024_30_jan_morning Work Done in Cyclic Process
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path Aarrow Barrow Carrow A as shown in the diagram. The total work done in the process is ________ J.
Work Done in Cyclic Process diagram for Q83 - JEE Main 2024 Morning
The image is a graph of Volume (dm3) vs Pressure (kPa) showing a triangular cyclic process starting from A(10,10) to B(10,30) to C(30,10) and back to A.
Numerical Answer. Answer: 200 to 200

Solution

Related Formula
Wcyclic = 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) on the y-axis versus Pressure (P) on the x-axis. The path A arrow B arrow C arrow A is traced in a clockwise direction on the V-P graph. Clockwise on a V-P graph corresponds to anti-clockwise on a standard P-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 kPa Height of triangle on V-axis = 30 - 10 = 20 dm³

Area = (1)/(2) × base × height Area = (1)/(2) × 20 × 20 = 200 kPa ³
Step 2: Unit conversion

1 kPa = 10³ Pa 1 dm³ = 1 Litre = 10⁻³ m³

W = 200 × 10³ Pa × 10⁻³ m³ W = 200 J
Pattern Recognition

1 kPa · 1 L = 1 Joule. This direct conversion saves time without converting explicitly to standard SI units (Pa and m³).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion
If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work w, is -x J. The value of x is ________ (Given R = 8.314 J K⁻¹mol⁻¹)
Numerical Answer. Answer: 28720 to 28721

Solution

Related Formula
W = -2.303 nRT ( (V₂)/(V₁) )
Core Logic

For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: n = 5 moles R = 8.314 J K⁻¹mol⁻¹ T = 300 K V₁ = 10 L V₂ = 100 L

Step 1: Calculating Work Done
W = -2.303 × 5 × 8.314 × 300 × ( (100)/(10) ) W = -2.303 × 5 × 8.314 × 300 × (10) W = -2.303 × 12471 × 1 W = -28720.713 J
Step 2: Final Formatting

The question asks for work w = -x J. So x = 28720.713, which rounds to 28721.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q88 jee_main_2024_31_jan_morning Gibbs Free Energy and Equilibrium
Consider the following reaction at 298 K. (3)/(2)O2(g) leftharpoons O3(g). Kₚ = 2.47 × 10⁻²⁹ ΔᵣG for the reaction is ________ kJ. (Given R = 8.314 J K⁻¹ mol⁻¹)
Numerical Answer. Answer: 163 to 164

Solution

Related Formula
ΔᵣG = -RT ln Kₚ
Step 1: Calculation
ΔᵣG = -8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ × 298 K × ln(2.47 × 10⁻²⁹) = -8.314 × 10⁻³ × 298 × (-65.87) = 163.19 kJ
Step 2: Nearest Integer

Rounding 163.19 to the nearest integer gives 163.

Chapter Mix

Class 11 Chemistry: Thermodynamics

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