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

Q jee_main_2025_29_jan_morning First Law of Thermodynamics and Heat Capacity

500 ~J

of energy is transferred as heat to

0.5 ~mol

of Argon gas at

298 ~K

and

1.00 atm

. The final temperature and the change in internal energy respectively are : Given:

R = 8.3 JK⁻¹ mol⁻¹

A. $348K$ $348\mathrm{K}$ and $300J$ $300\mathrm{J}$
B. $378K$ $378\mathrm{K}$ and $300J$ $300\mathrm{J}$
C. $368K$ $368\mathrm{K}$ and $500J$ $500\mathrm{J}$
D. $378K$ $378\mathrm{K}$ and $500J$ $500\mathrm{J}$

Solution

Formulas Used

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

qₚ = n \cdot Cₚ \cdot Δ T

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

Δ U = n \cdot Cv \cdot Δ T

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 ( $qₚ$ $q_p$) = $500 J$ $500\text{ J}$

500 = 0.5 \times ((5)/(2) \times 8.3) \times (Tf - 298)

500 = 0.5 \times 20.75 \times (Tf - 298)

500 = 10.375 \times (Tf - 298)

Tf - 298 = (500)/(10.375) \approx 48.2 K

Tf = 298 + 48.2 = 346.2 K \approx 348 K

---

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

Δ U = n \cdot Cv \cdot Δ T

Alternatively, using the ratio of heat capacities:

Δ U = ((Cv)/(Cₚ)) \times qₚ = (3)/(5) \times 500 J = 300 J

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)

Q79 jee_main_2024_01_february_morning First Law of Thermodynamics

Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:

A. $q = 0, Δ T \neq 0, w = 0$ $q = 0, \Delta T \neq 0, w = 0$
B. $q = 0, Δ T < 0, w \neq 0$ $q = 0, \Delta T < 0, w \neq 0$
C. $q \neq 0, Δ T = 0, w = 0$ $q \neq 0, \Delta T = 0, w = 0$
D. $q = 0, Δ T = 0, w = 0$ $q = 0, \Delta T = 0, w = 0$

Solution

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. $q = 0, w = 0, Δ U = 0, Δ T = 0, Δ H = 0$ .

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q jee_main_2024_29_january_evening Enthalpy of Phase Transition

Standard enthalpy of vapourisation for

CCl₄

is

30.5 kJ mol⁻¹

. Heat required for vapourisation of

284g

of

CCl₄

at constant temperature is ________ kJ. (Given molar mass in g mol-; C = 12, Cl = 35.5)

Numerical Answer. Answer: 56 to 56.25

Solution

Related Formula

Q = n \times Δ Hvap⁰ where n = MassMolar Mass

Core Logic

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

Molar mass = 12 + 4(35.5) = 12 + 142 = 154 g/mol

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

n = (284)/(154) \approx 1.844 moles

Step 1: Enthalpy Calculation

Calculate the total energy required for vaporization:

Δ H = 1.844 mol \times 30.5 kJ/mol \approx 56.24 kJ

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

Q84 jee_main_2024_27_jan_morning Isothermal Expansion and Work Calculations

If three moles of an ideal gas at

300 K

expand isothermally from

30 dm³

to

45 dm³

against a constant opposing pressure of

80 kPa

, then the amount of heat transferred is

J.

Numerical Answer. Answer: 1200 to 1200

Solution

Related Formula

First law of thermodynamics framework:

Δ U = Q + W

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

Δ U = 0 Q = -W

Irreversible work formula expanding against constant external pressure:

W = -Pₑₓₜ Δ V = -Pₑₓₜ(V₂ - V₁)

Step 1: Calculate structural work values

Given values:

Pₑₓₜ = 80 kPa = 80 \times 10³ Pa

V₁ = 30 dm³ = 30 \times 10⁻³ m³

V₂ = 45 dm³ = 45 \times 10⁻³ m³

Δ V = (45 - 30) \times 10⁻³ = 15 \times 10⁻³ m³

W = -80 \times 10³ \times (15 \times 10⁻³) = -1200 J

Step 2: Solve for heat magnitude

$Q = -W = -(-1200 J) = 1200 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

Q jee_main_2024_29_jan_morning Spontaneity and Gibbs Free Energy

Which of the following is not correct?

A. $Δ G is negative for a spontaneous reaction$
B. $Δ G is positive for a spontaneous reaction$
C. $Δ G is zero for a reversible reaction$
D. $Δ G is positive for a non-spontaneous reaction$

Solution

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

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 8

Solution

Formulas Used

Core Logic

Pattern Recognition

Solution

Core Logic

Step 1: Final Parameter Check

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Enthalpy Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Calculate structural work values

Step 2: Solve for heat magnitude

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

More Thermodynamics Questions — jee_main_2025_03_april_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

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).