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 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

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