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 4

Q48 jee_main_2025_03_april_evening Bomb Calorimetry and Heat of Combustion

A sample of n-octane (

1.14~g

) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is

5 ~kJ~K⁻¹

. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is ________

kJ~mol⁻¹

. (nearest integer)

Numerical Answer. Answer: 2500 to 2500

Solution

Related Formula

Heat released at constant volume ( $qv$ $q_v$) in a bomb calorimeter is:

qv = Ccal \cdot Δ T

Molar heat of combustion at constant volume ( $Δ Ucomb$ $\Delta U_{\text{comb}}$):

Δ Ucomb = qvnfuel

Core Logic

Given parameters:

Mass of n-octane $m = 1.14~g$ $m = 1.14\mathrm{~g}$
Heat capacity of calorimeter $Ccal = 5~kJ/K$ $C_{\text{cal}} = 5\mathrm{~kJ/K}$
Temperature rise $Δ T = 5~K$ $\Delta T = 5\mathrm{~K}$
Formula of octane: $C₈H₁₈$ $\mathrm{C_8H_{18}}$ $\Rightarrow$ $\Rightarrow$ Molar mass $= 8(12) + 18(1) = 114~g/mol$ $= 8(12) + 18(1) = 114\mathrm{~g/mol}$

Step 1: Calculate heat absorbed by the calorimeter ( $qv$ $q_v$)

qv = 5~kJ/K \times 5~K = 25~kJ

Step 2: Calculate moles of octane

n = 1.14~g114~g/mol = 0.01~mol

Step 3: Calculate molar heat of combustion

Δ Ucomb = 25~kJ0.01~mol = 2500~kJ/mol

The magnitude of the heat of combustion is$ The magnitude of the heat of combustion is $2500\mathrm{~kJ~mol^{-1}}. Pattern Recognition A bomb calorimeter operates at rigid constant volume, meaning boundary work$. Pattern Recognition A bomb calorimeter operates at rigid constant volume, meaning boundary work $w = 0. Thus, by the first law of thermodynamics, the measured heat flow represents the internal energy change ($. Thus, by the first law of thermodynamics, the measured heat flow represents the internal energy change ($\Delta U), not the enthalpy change ($), not the enthalpy change ($\Delta H$, which occurs at constant pressure). Chapter Mix Class 11 Chemistry: Thermodynamics

Q29 jee_main_2025_07_april_morning Enthalpy Changes

Total enthalpy change for freezing of

1 ~mol

of water at

10° C

to ice at

-10° C

is (Given:

ΔfusH = x

kJ/mol,

Cₚ[H₂O( )] = y J mol⁻¹ K⁻¹

Cₚ[H₂O(s)] = z J mol⁻¹ K⁻¹

)

A. $-x - 10y - 10z$
B. $-10(100x + y + z)$
C. $10(100x + y + z)$
D. $x - 10y - 10z$

Solution

Related Formula

Δ Htotal = n Cp( ) Δ T₁ - nΔ Hfusion + n Cp(s) Δ T₂

Core Logic

We need to compute the enthalpy change for the pathway:

H₂O( , 10°C) arrow H₂O(s, -10°C)

This can be divided into three consecutive steps:

Cool liquid water from $10°C$ $10^{\circ}C$ to $0°C$ $0^{\circ}C$:

Δ H₁ = n \cdot Cₚ[H₂O( )] \cdot (0 - 10) = 1 \cdot y \cdot (-10) = -10y J

Freeze water to ice at $0°C$ $0^{\circ}C$:

Δ H₂ = -n \cdot ΔfusH = -1 \cdot x kJ = -1000x J

Cool ice from $0°C$ $0^{\circ}C$ to $-10°C$ $-10^{\circ}C$:

Δ H₃ = n \cdot Cₚ[H₂O(s)] \cdot (-10 - 0) = 1 \cdot z \cdot (-10) = -10z J

Thermodynamics enthalpy cycle for freezing Q29

Adding these three steps yields:

Δ Htotal = -10y - 1000x - 10z = -10(100x + y + z) Joule

Pattern Recognition

Freezing is an exothermic process, so all three steps (cooling water, freezing, and cooling ice) must carry a negative sign. Factoring out $-10$ $-10$ cleanly yields the expression $-10(100x+y+z)$ $-10(100x+y+z)$.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q jee_main_2025_08_april_evening Resonance Energy

Resonance in an

X₂Y

molecule is represented as follows:

X = X = Y X \equiv X^+ - Y^-

The experimental enthalpy of formation for gaseous

X₂Y

is given by the reaction:

X \equiv X(g) + (1)/(2)Y = Y(g) X₂Y(g) Δ Hf(exp) = 80 kJ mol⁻¹

Calculate the magnitude of the resonance energy of

X₂Y

kJ mol⁻¹

(as the nearest integer value). **Given bond energies:** *

X \equiv X = 940 kJ mol⁻¹

X = X = 410 kJ mol⁻¹

Y = Y = 500 kJ mol⁻¹

X = Y = 602 kJ mol⁻¹

Valence settings:

X : 3, Y : 2

Numerical Answer. Answer: 98 to 98

Solution

Related Formula

Resonance energy equation related to experimental and theoretical formation enthalpies:

Δ HR.E. = Δ Hf(exp) - Δ Hf(Theo)

Theoretical enthalpy calculation using bond energies:

Δ Hf(Theo) = Σ B.E.reactants - Σ B.E.products

Execution

Step 1: Write the chemical equation to calculate the theoretical enthalpy of formation based on the localized structure $X=X=Y$ $X=X=Y$:

X \equiv X(g) + (1)/(2)Y=Y(g) X=X=Y(g)

Step 2: Substitute the localized bond energies into the reactant-minus-product relation:

Δ Hf(Theo) = [ B.E.X \equiv X + (1)/(2)B.E.Y=Y ] - [ B.E.X=X + B.E.X=Y ]

Δ Hf(Theo) = [ 940 + (1)/(2)(500) ] - [410 + 602]

Δ Hf(Theo) = [940 + 250] - 1012 = 1190 - 1012 = 178 kJ mol⁻¹

Step 3: Calculate the resonance energy:

Δ HR.E. = Δ Hf(exp) - Δ Hf(Theo) = 80 - 178 = -98 kJ mol⁻¹

Taking the magnitude as requested: $|Δ HR.E.| = 98$ $|\Delta H_{\text{R.E.}}| = 98$.

Pattern Recognition

Resonance energy is always a stabilizing factor, meaning $Δ Hf(exp)$ $\Delta H_{f(\text{exp})}$ is more exothermic (or less endothermic) than the theoretical localized state. The magnitude is simply the absolute value of this difference ( $|80 - 178| = 98$ $|80 - 178| = 98$).

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q39 jee_main_2025_29_jan_evening Hess's Law of Constant Heat Summation

C(diamond) arrow C(graphite) + X kJ mol⁻¹

C(diamond) + O₂(g) arrow CO₂(g) + Y kJ mol⁻¹

C(graphite) + O₂(g) arrow CO₂(g) + Z kJ mol⁻¹

At constant temperature, then the correct relationship is:

A. X = Y + Z
B. -X = Y + Z
C. X = -Y + Z
D. X = Y - Z

Solution

Core Logic

Let's treat the given parameters as terms for exothermic heats evolved on the product side: 1) $C(diamond) arrow C(graphite)$ , $Delta H = -X$ $Delta H = -X$ 2) $C(diamond) + O₂(g) arrow CO₂(g)$ , $Delta H = -Y$ $Delta H = -Y$ 3) $C(graphite) + O₂(g) arrow CO₂(g)$ , $Delta H = -Z$ $Delta H = -Z$

By subtracting Equation (3) from Equation (2):

C(diamond) - C(graphite) arrow 0 implies C(diamond) arrow C(graphite)

Delta H = (-Y) - (-Z) = Z - Y

Matching this to Equation (1):

-X = Z - Y implies X = Y - Z

Pattern Recognition

Hess's Law states that the enthalpy change of an overall reaction is equal to the sum of the enthalpy changes of its individual steps.

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 4

Solution

Related Formula

Core Logic

Step 1: Calculate heat absorbed by the calorimeter (qv$q_v$)

Step 2: Calculate moles of octane

Step 3: Calculate molar heat of combustion

Step 3: Calculate molar heat of combustion

Pattern Recognition

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Execution

Pattern Recognition

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

Step 1: Calculate heat absorbed by the calorimeter ( $qv$ $q_v$)