Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

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: 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: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the melting point decreases, causing the entire solid phase (ice) to disappear completely. 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. Class 11 Chemistry: Chemical Thermodynamics

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct obse

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: 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: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the meltin

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

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: 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: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the melting point decreases, causing the entire solid phase (ice) to disappear completely. 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. Class 11 Chemistry: Chemical Thermodynamics

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct obse

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: 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: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the meltin

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the…

Q47 jee_main_2025_28_jan_evening Hess's Law / Enthalpy of Formation

Consider the following data: Heat of formation of

CO_2(g) = -393.5mathrm~kJ~mol^-1

Heat of formation of

H_2O(l) = -286.0mathrm~kJ~mol^-1

Heat of combustion of benzene

= -3267.0mathrm~kJ~mol^-1

The heat of formation of benzene is ______

mathrmkJ~mol^-1

(Nearest integer).

Numerical Answer. Answer: 48 to 48

Solution

### Related Formula Enthalpy of reaction from enthalpy of formation data:

Delta H_textreaction = sum Delta H_f(textProducts) - sum Delta H_f(textReactants)

### Core Logic Write out the balanced thermochemical equation for the combustion of benzene (

C_6H_6

):

C_6H_6(l) + frac152O_2(g) rightarrow 6CO_2(g) + 3H_2O(l)

Given parameters: -

Delta H_c = -3267.0mathrm\ kJ/mol

-

Delta H_f[CO_2] = -393.5mathrm\ kJ/mol

-

Delta H_f[H_2O] = -286.0mathrm\ kJ/mol

-

Delta H_f[O_2] = 0mathrm\ kJ/mol

### Step 1: Applying Hess's Law Substitute these values into the reaction expression:

-3267 = [6(-393.5) + 3(-286.0)] - Delta H_f[C_6H_6]

-3267 = [-2361.0 - 858.0] - Delta H_f[C_6H_6]

-3267 = -3219.0 - Delta H_f[C_6H_6]

Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48mathrm\ kJ/mol

### Pattern Recognition Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply

CO_2

by

6

and

H_2O

by

3

) to ensure accurate bookkeeping. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

Q jee_main_2025_29_jan_morning First Law of Thermodynamics and Heat Capacity

500 mathrm~J

of energy is transferred as heat to

0.5 mathrm~mol

of Argon gas at

298 mathrm~K

and

1.00 mathrmatm

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

mathrmR = 8.3 \, mathrmJK^-1 mathrmmol^-1

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

Solution

### Formulas Used For an ideal gas undergoing a constant pressure process (

1.00text atm

):

q_p = n cdot C_p cdot Delta T

Change in internal energy (

Delta U

):

Delta U = n cdot C_v cdot Delta T

For a monoatomic gas like Argon: *

C_v = frac32 R

*

C_p = frac52 R

### Core Logic **Step 1: Calculate the final temperature (

T_f

)** Heat transferred at constant pressure (

q_p

) =

500text J

500 = 0.5 times left(frac52 times 8.3right) times (T_f - 298)

500 = 0.5 times 20.75 times (T_f - 298)

500 = 10.375 times (T_f - 298)

T_f - 298 = frac50010.375 approx 48.2text K

T_f = 298 + 48.2 = 346.2text K approx 348text K

--- **Step 2: Calculate the change in internal energy (

Delta U

)**

Delta U = n cdot C_v cdot Delta T

Alternatively, using the ratio of heat capacities:

Delta U = left(fracC_vC_pright) times q_p = frac35 times 500text J = 300text J

Thus, the final temperature is **

348text K

** and the change in internal energy is **

300text J

**. ### Pattern Recognition For a monoatomic ideal gas under constant pressure, exactly

60\%

of the heat added (

fracC_vC_p = frac35

) goes into increasing the internal energy (

Delta U

), while

40\%

is lost to expansion work (

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

Solution

### Core Logic Free expansion means expansion against a vacuum (

P_ext = 0

). Work done:

w = -P_ext Delta V

. Since

P_ext = 0

,

w = 0

. Adiabatic condition means there is no heat exchange with the surroundings. Heat transfer:

q = 0

. According to the First Law of Thermodynamics,

Delta U = q + w

. Since

q = 0

and

w = 0

, the change in internal energy

Delta U = 0

. For an ideal gas, internal energy is a function of temperature only (

Delta U = nC_vDelta T

). If

Delta U = 0

, then

Delta T = 0

. ### Step 1: Final Parameter Check Evaluating all parameters simultaneously:

q = 0

w = 0

Delta T = 0

### Pattern Recognition Adiabatic + Free Expansion of IDEAL gas

rightarrow

Nothing changes thermodynamically except volume and pressure.

q = 0, w = 0, Delta U = 0, Delta T = 0, Delta H = 0

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

Q88 jee_main_2024_29_january_evening Enthalpy of Phase Transition

Standard enthalpy of vapourisation for

mathrmCCl_4

is

30.5mathrmkJ mol^-1

. Heat required for vapourisation of

284mathrmg

of

mathrmCCl_4

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 Delta H_textvap^0 quad textwhere n = fractextMass, textMolar Mass

### Core Logic First, calculate the molar mass of carbon tetrachloride (

CCl_4

):

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

Next, calculate the total number of moles present in

284text g

of the substance:

n = frac284, 154 approx 1.844text moles

### Step 1: Enthalpy Calculation Calculate the total energy required for vaporization:

Delta H = 1.844text mol times 30.5text kJ/mol approx 56.24text kJ

Rounding to the nearest integer value gives **56**. ### Pattern Recognition Enthalpy of vaporization is an intensive property given per mole. Scale it lineary by multiplying by the total number of moles to find the total extensive heat required. ### Evaluation Rubric / Model Answer null ### 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

300text K

expand isothermally from

30text dm^3

to

45text dm^3

against a constant opposing pressure of

80text kPa

, then the amount of heat transferred is

textquadquad

J.

Numerical Answer. Answer: 1200 to 1200

Solution

### Related Formula First law of thermodynamics framework:

Delta U = Q + W

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

Delta U = 0 implies Q = -W

Irreversible work formula expanding against constant external pressure:

W = -P_textext Delta V = -P_textext(V_2 - V_1)

### Step 1: Calculate structural work values Given values:

P_textext = 80text kPa = 80 times 10^3text Pa

V_1 = 30text dm^3 = 30 times 10^-3text m^3

V_2 = 45text dm^3 = 45 times 10^-3text m^3

Delta V = (45 - 30) times 10^-3 = 15 times 10^-3text m^3

W = -80 times 10^3 times (15 times 10^-3) = -1200text J

### Step 2: Solve for heat magnitude

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

### 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

Ice and water are placed in a closed container at a pressure of 1 atm and temperature $273.15mathrmK$ $273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

Solution & Explanation

More Chemical Thermodynamics Previous-Year Questions — Page 6

Solution

Solution

Solution

Solution

Solution

More Chemical Thermodynamics Questions — jee_main_2025_28_jan_morning

We Map Every Repeating Question in Competitive Exams.

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

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15mathrmK$273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

Solution & Explanation

More Chemical Thermodynamics Previous-Year Questions — Page 6

Solution

Solution

Solution

Solution

Solution

More Chemical Thermodynamics Questions — jee_main_2025_28_jan_morning

We Map Every Repeating Question in Competitive Exams.

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

Ice and water are placed in a closed container at a pressure of 1 atm and temperature $273.15mathrmK$ $273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following: