A perfect gas (0.1~mol) having Cv=1.50~R (independent of temperature) undergoes the transformation from point 1 to…

Year	2026	2025	2024	Total
Questions	12	24	8	44

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Thermodynamic work done ( $w$ $w$) for each step:

Isochoric step ( $V = constant$ $V = \text{constant}$):

$w = 0$ $w = 0$

Isobaric step ( $P = constant$ $P = \text{constant}$):

$w = -P Δ V$ $w = -P \Delta V$

Core Logic

The process from point 1 to point 4 consists of three distinct segments:

Step $1 arrow 2$ $1 \rightarrow 2$: Isochoric cooling at constant volume $V₁ = 1000~cm³$ $V_1 = 1000\mathrm{~cm}^3$. Work done $w1arrow 2 = 0$ $w_{1\rightarrow 2} = 0$.
Step $2 arrow 3$ $2 \rightarrow 3$: Isobaric compression at constant pressure $P = 3.00~atm$ $P = 3.00\mathrm{~atm}$ from volume $2000~cm³$ $2000\mathrm{~cm}^3$ to $1000~cm³$ $1000\mathrm{~cm}^3$.
Step $3 arrow 4$ $3 \rightarrow 4$: Isochoric step at constant volume. Work done $w3arrow 4 = 0$ $w_{3\rightarrow 4} = 0$.

Step 1: Calculate work done in the isobaric step ( $2 arrow 3$ $2 \rightarrow 3$)

The volume changes from $Vᵢ = 2000~cm³ = 2.0~L$ to $Vf = 1000~cm³ = 1.0~L$ :

w2arrow 3 = -P Δ V = -3.00~atm \times (1.0~L - 2.0~L) = +3.00~L\cdot atm

Step 2: Convert work to Joules and analyze direction Convert$ Step 2: Convert work to Joules and analyze direction Convert $\mathrm{L\cdot atm}to Joules:

w2arrow 3 = 3.00 \times 101.325~J = 303.975~J \approx 304~J

The question asks for the total work done as ($ The question asks for the total work done as ($-) ________ J, meaning work done by the system (expansion) is negative and work done on the system (compression) is positive. Since this is compression, work done on the gas is$) ________ J, meaning work done by the system (expansion) is negative and work done on the system (compression) is positive. Since this is compression, work done on the gas is $+304\mathrm{~J}, which is represented as$, which is represented as $-(-304)\mathrm{~J}in typical IUPAC convention where work of expansion is examined. The absolute magnitude of the work is$ in typical IUPAC convention where work of expansion is examined. The absolute magnitude of the work is $304\mathrm{~J}. Pattern Recognition During any cyclic or multi-step path on a P-V graph, work is done only when there is a change in volume ($. Pattern Recognition During any cyclic or multi-step path on a P-V graph, work is done only when there is a change in volume ($W = -\int P dV$). Any vertical line (constant volume) represents an isochoric step where work is exactly zero. The horizontal segment directly represents rectangular area under the path. Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermodynamics

More Thermodynamics Previous-Year Questions — Page 8

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

30.5 kJ mol⁻¹

. Heat required for vapourisation of

284g

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

45 dm³

against a constant opposing pressure of

80 kPa

, then the amount of heat transferred is

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate work done in the isobaric step (2 arrow 3$2 \rightarrow 3$)

Step 2: Convert work to Joules and analyze direction

Step 2: Convert work to Joules and analyze direction

Pattern Recognition

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

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Step 1: Calculate work done in the isobaric step ( $2 arrow 3$ $2 \rightarrow 3$)