Which of the following graphs between pressure ‘P’ versus volume ‘V’ represent the maximum work done?

The magnitude of work done by or on a gas is given by the area under the P-V curve projected onto the volume axis. Graph 1: Cyclic process forming a triangle. Area is bounded, represents net work. Graph 2: Isochoric drop (vertical line at V=22.4L). Area = 0, so work done is zero. Graph 3: Expansion process forming a cycle. Area is enclosed in a convex shape. Graph 4: Direct expansion from V=22.4 to V=44.8 at pressure P=1 up to P=2 (a rectangle combined with a triangle). The total area under the upper curve from V=22.4 to V=44.8 covers the entire shaded region under the path down to the V-axis. Option (4) provides the largest total area under the curve extending down to the horizontal axis (maximum magnitude of work done). Class 11 Chemistry: Thermodynamics

Which of the following graphs between pressure ‘P’ versus volume ‘V’ represent the maximum work done?

The magnitude of work done by or on a gas is given by the area under the P-V curve projected onto the volume axis. Graph 1: Cyclic process forming a triangle. Area is bounded, represents net work. Graph 2: Isochoric drop (vertical line at V=22.4L). Area = 0, so work done is zero. Graph 3: Expansion process forming a cycle. Area is enclosed in a convex shape. Graph 4: Direct expansion from V=22.4 to V=44.8 at pressure P=1 up to P=2 (a rectangle combined with a triangle). The total area under the upper curve from V=22.4 to V=44.8 covers the entire shaded region under the path down to the V-axis. Option (4) provides the largest total area under the curve extending down to the horizontal axis (maximum magnitude of work done). Class 11 Chemistry: Thermodynamics

Solution & Explanation

### Core Logic The magnitude of work done by or on a gas is given by the area under the P-V curve projected onto the volume axis. Graph 1: Cyclic process forming a triangle. Area is bounded, represents net work. Graph 2: Isochoric drop (vertical line at

V=22.4L

). Area = 0, so work done is zero. Graph 3: Expansion process forming a cycle. Area is enclosed in a convex shape. Graph 4: Direct expansion from

V=22.4

V=44.8

at pressure

P=1

up to

P=2

(a rectangle combined with a triangle). The total area under the upper curve from

V=22.4

V=44.8

covers the entire shaded region under the path down to the V-axis. Option (4) provides the largest total area under the curve extending down to the horizontal axis (maximum magnitude of work done). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

More Thermodynamics Previous-Year Questions — Page 3

Q39 jee_main_2025_29_jan_evening Hess's Law of Constant Heat Summation

C(textdiamond) ightarrow C(textgraphite) + Xtext kJ mol^-1

C(textdiamond) + O_2(g) ightarrow CO_2(g) + Ytext kJ mol^-1

C(textgraphite) + O_2(g) ightarrow CO_2(g) + Ztext 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(textdiamond) ightarrow C(textgraphite)

Delta H = -X

C(textdiamond) + O_2(g) ightarrow CO_2(g)

Delta H = -Y

C(textgraphite) + O_2(g) ightarrow CO_2(g)

Delta H = -Z

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

C(textdiamond) - C(textgraphite) ightarrow 0 implies C(textdiamond) ightarrow C(textgraphite)

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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

Q34 jee_main_2025_28_jan_morning Phase Equilibrium and Le Chatelier's Principle

Ice and water are placed in a closed container at a pressure of 1 atm and temperature

273.15mathrmK

. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

A. $textVolume of system increases.$ $\text{Volume of system increases.}$
B. $textLiquid phase disappears completely.$
C. $textThe amount of ice decreases.$ $\text{The amount of ice decreases.}$
D. $textThe solid phase (ice) disappears completely.$

Solution

### Core Logic Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water:

V_m(textice) > V_m(textwater)

According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water:

Phase shift diagram for Q34 - JEE Main 2025 Morning

If the pressure is increased considerably (such as doubling it to 2 atm) at

273.15mathrmK

, the melting point decreases, causing the entire solid phase (ice) to disappear completely. ### Pattern Recognition Sees: Ice-water system under pressure change. Trap: Assuming that an increase in pressure always favors the solid phase. Water has an anomalous phase curve with a negative slope. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

Q50 jee_main_2025_28_jan_morning Bond Enthalpy Calculation

The formation enthalpies,

Delta mathrmH_mathrmf^ominus

for

mathrmH_(mathrmg)

and

mathrmO_(mathrmg)

are 220.0 and

250.0~mathrmkJ~mol^-1

, respectively, at

298.15mathrmK

, and

Delta mathrmH_mathrmf^-

for

mathrmH_2mathrmO_(mathrmg)

-242.0mathrmkJ\,mol^-1

at the same temperature. The average bond enthalpy of the O-H bond in water at

298.15mathrmK

\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (nearest integer).

Numerical Answer. Answer: 466 to 466

Solution

### Related Formula Reaction enthalpy based on atomization processes:

Delta_r H = sum Delta_f H(textproducts) - sum Delta_f H(textreactants)

### Step 1: Map the Dissociation Reaction Consider the dissociation of gas phase water molecules into constituent gaseous atoms:

mathrmH_2mathrmO_mathrm(g) rightarrow 2mathrmH_mathrm(g) + mathrmO_mathrm(g)

The total energy required corresponds to breaking exactly two

mathrmO-mathrmH

bonds:

Delta_r H = 2 times textB.E.(mathrmO-mathrmH)

### Step 2: Calculate

Delta_r H

Using the enthalpies of formation:

Delta_r H = [2 times Delta_f H(mathrmH_mathrm(g)) + Delta_f H(mathrmO_mathrm(g))] - Delta_f H(mathrmH_2mathrmO_mathrm(g))

Delta_r H = [2 times 220.0 + 250.0] - (-242.0)

Delta_r H = [440.0 + 250.0] + 242.0 = 690.0 + 242.0 = 932.0\,mathrmkJ\,mol^-1

### Step 3: Solve for Single Bond Enthalpy

2 times textB.E.(mathrmO-mathrmH) = 932.0

textB.E.(mathrmO-mathrmH) = frac932.02 = 466\,mathrmkJ\,mol^-1

### Pattern Recognition Sees: Atomization state values used to evaluate single bond metrics. Shortcut: Remember

textTotal Dissociation Energy = sum Delta_f H(textatoms) - Delta_f H(textmolecule)

. Halving the result gives the average bond enthalpy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

Q35 jee_main_2025_03_april_morning Intensive and Extensive Properties

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

A. Molar heat capacity
B. Density
C. Concentration
D. Gibbs free energy

Solution

### Core Logic Let us compute the concentration of both solutions:

textConcentration of Solution 1 = frac10text mol10text L = 1text mol/L

textConcentration of Solution 2 = frac1text mol1text L = 1text mol/L

Since concentration is identical, both systems share matching compositions. Consequently, all **intensive properties** (independent of mass/size) like concentration, density, and molar heat capacity remain exactly equal. ### Step 1: Identifying the Variable Gibbs free energy (

G

) is an **extensive property** that scales directly with the amount of matter in the system. Because Solution 1 contains a larger total mass and volume than Solution 2, its overall Gibbs free energy value is different. ### Pattern Recognition Shortcut: Look for the only extensive property in the options. Density, concentration, and molar parameters are always intensive. Gibbs free energy (

G

) scales with total matter quantity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

Q49 jee_main_2025_03_april_morning Bond Enthalpy and Enthalpy of Formation

Given:

Delta Htextsub^ominus[C(textgraphite)] = 710text kJ mol^-1

DeltaC-HH^ominus = 414text kJ mol^-1

DeltaH-HH^ominus = 436text kJ mol^-1

DeltaC=CH^ominus = 611text kJ mol^-1

The

Delta Hf^ominus

for

CH2=CH2

is ________

textkJ mol^-1

(nearest integer value)

Numerical Answer. Answer: 25 to 25

Solution

### Related Formula The enthalpy of formation reaction can be evaluated by balancing atomization and bond cleavage details:

Delta H_f^circ = sum Delta Htextatomization (reactants) - sum Delta Htextbonds broken/formed (products)

### Core Logic The target formation reaction for ethylene (

textC_2textH_4

) from standard elemental states is:

2C(textgraphite) + 2H_2(g) ightarrow CH_2=CH_2(g)

To construct this pathway: 1. Sublime 2 moles of solid graphite:

2 times Delta H_textsub^circ[C]

2. Dissociate 2 moles of gaseous

textH-textH

bonds:

2 times Delta_H-HH^circ

3. Form 1 mole of

textC=textC

double bonds:

-1 times Delta_C=CH^circ

4. Form 4 moles of

textC-textH

single bonds:

-4 times Delta_C-HH^circ

### Step 1: Arithmetic Calculation

Delta H_f^circ = [2 times 710] + [2 times 436] - 611 - [4 times 414]

Delta H_f^circ = 1420 + 872 - 611 - 1656 = 2292 - 2267 = 25text kJ mol^-1

### Pattern Recognition Shortcut: Group energy components systematically. Reactant state atomization costs

+2292text kJ

. Exothermic structural bond formation releases

-2267text kJ

. Net difference yields a small endothermic value of

+25text kJ/mol

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

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