Which of the following properties will change when system containing solution 1 will become solution 2?

Q: Which of the following properties will change when system containing solution 1 will become solution 2?

G = H - TS Extensive properties depend on mass/amount of matter; Intensive properties are independent of amount. Both solution 1 (10 mol/10 L = 1 M) and solution 2 (1 mol/1 L = 1 M) have identical concentrations and chemical compositions. Intensive properties (molar heat capacity, density, concentration) depend only on composition and temperature, so they remain unchanged. Gibbs free energy (G) is an extensive property, proportional to the total amount of substance present, so it changes. Gibbs

Year	2026	2025	2024	Total
Questions	12	24	8	44

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

$G = H - TS$ $G = H - TS$

Extensive properties depend on mass/amount of matter; Intensive properties are independent of amount.

Core Logic

Both solution 1 ( $10 mol/10 L = 1 M$ ) and solution 2 ( $1 mol/1 L = 1 M$ $1\text{ mol}/1\text{ L} = 1\text{ M}$) have identical concentrations and chemical compositions.

Intensive properties (molar heat capacity, density, concentration) depend only on composition and temperature, so they remain unchanged.

Gibbs free energy ( $G$ $G$) is an extensive property, proportional to the total amount of substance present, so it changes.

Step 1: Final Conclusion

Gibbs free energy changes because it is an extensive property.

Pattern Recognition

Same concentration/composition $arrow$ $\rightarrow$ Intensive properties stay equal. Total amount changes $arrow$ $\rightarrow$ Extensive properties ( $G, H, U, S$ $G, H, U, S$) change.

Chapter Mix

Class 11 Chemistry: Thermodynamics

More 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 $3 CO(g)$ $3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3.
We need $Fe₂O3(s)$ $\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and $2 Fe(s)$ $2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1).

Step 1: Calculate Net Enthalpy

Target Reaction (3) = $3 \times (2) - (1)$ $3 \times (2) - (1)$

Δ H₃ = 3 \times 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 ( $V$ $V$) on the y-axis versus Pressure ( $P$ $P$) on the x-axis. The path $A arrow B arrow C arrow A$ 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 = $30 - 10 = 20 kPa$ $30 - 10 = 20 \text{ kPa}$ Height of triangle on V-axis = $30 - 10 = 20 dm³$ $30 - 10 = 20 \text{ dm}^3$

Area = (1)/(2) \times base \times height

Area = (1)/(2) \times 20 \times 20 = 200 kPa ³

Step 2: Unit conversion

$1 kPa = 10³ Pa$ $1 \text{ kPa} = 10^3 \text{ Pa}$ $1 dm³ = 1 Litre = 10⁻³ m³$

W = 200 \times 10³ Pa \times 10⁻³ m³

W = 200 J

Pattern Recognition

$1 kPa \cdot 1 L = 1 Joule$ . This direct conversion saves time without converting explicitly to standard SI units ( $Pa$ $Pa$ and $m³$ $m^3$).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion

5 moles

of an ideal gas expands from

10 L

to a volume of

100 L

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: $n = 5 moles$ $n = 5\text{ moles}$ $R = 8.314 J K⁻¹mol⁻¹$ $T = 300 K$ $T = 300\text{ K}$ $V₁ = 10 L$ $V_1 = 10\text{ L}$ $V₂ = 100 L$ $V_2 = 100\text{ L}$

Step 1: Calculating Work Done

W = -2.303 \times 5 \times 8.314 \times 300 \times ( (100)/(10) )

W = -2.303 \times 5 \times 8.314 \times 300 \times (10)

W = -2.303 \times 12471 \times 1

W = -28720.713 J

Step 2: Final Formatting

The question asks for work $w = -x J$ $w = -x\text{ J}$. So $x = 28720.713$ $x = 28720.713$, which rounds to $28721$ $28721$.

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 \times 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 \times 10⁻³ kJ K⁻¹ mol⁻¹ \times 298 K \times ln(2.47 \times 10⁻²⁹)

= -8.314 \times 10⁻³ \times 298 \times (-65.87)

= 163.19 kJ

Step 2: Nearest Integer

Rounding $163.19$ $163.19$ to the nearest integer gives $163$ $163$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Which of the following properties will change when system containing solution 1 will become solution 2? Diagram showing solution 1 (10 mol solute in 10 L water) and solution 2 (1 mol solute in 1 L water).

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Calculate Net Enthalpy

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Area

Step 2: Unit conversion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Work Done

Step 2: Final Formatting

Chapter Mix

Solution

Related Formula

Step 1: Calculation

Step 2: Nearest Integer

Chapter Mix

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Which of the following properties will change when system containing solution 1 will become solution 2?
Diagram showing solution 1 (10 mol solute in 10 L water) and solution 2 (1 mol solute in 1 L water).