Match List-I with List-II. List-I List-II (A) Isothermal (I) Delta W (work done) =0 (B) Adiabatic (II) Delta Q (supplied heat) =0 (C) Isobaric (III) Delta U (change in internal energy) 0 (D) Isochoric (IV) Delta U=0 Choose the correct answer from the options given below: [cite: 151, 152]

Let's evaluate each process condition based on the first law of thermodynamics: * **(A) Isothermal:** Continuous constant temperature (Delta T = 0) implies that the internal energy change of an ideal gas is zero, so Delta U = 0 (IV) [cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 (II) [cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, Delta U neq 0 (III) [cite: 732]. * **(D) Isochoric:** Rigid boundary condition at constant volume (Delta V = 0) ensures work done Delta W = PDelta V = 0 (I) [cite: 733]. Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729]. Matching 'Isochoric' with zero work done (Delta W=0) or 'Adiabatic' with zero heat exchange (Delta Q=0) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733]. Class 11 Physics: Thermodynamics

Match List-I with List-II. List-I List-II (A) Isothermal (I) Delta W (work done) =0 (B) Adiabatic (II) Delta Q (supplied heat) =0 (C) Isobaric (III) Delta U (change in internal energy) 0 (D) Isochoric

Let's evaluate each process condition based on the first law of thermodynamics: * **(A) Isothermal:** Continuous constant temperature (Delta T = 0) implies that the internal energy change of an ideal gas is zero, so Delta U = 0 (IV) [cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 (II) [cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes c

Match List-I with List-II. List-I List-II (A) Isothermal (I) Delta W (work done) =0 (B) Adiabatic (II) Delta Q (supplied heat) =0 (C) Isobaric (III) Delta U (change in internal energy) 0 (D) Isochoric (IV) Delta U=0 Choose the correct answer from the options given below: [cite: 151, 152]

Let's evaluate each process condition based on the first law of thermodynamics: * **(A) Isothermal:** Continuous constant temperature (Delta T = 0) implies that the internal energy change of an ideal gas is zero, so Delta U = 0 (IV) [cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 (II) [cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, Delta U neq 0 (III) [cite: 732]. * **(D) Isochoric:** Rigid boundary condition at constant volume (Delta V = 0) ensures work done Delta W = PDelta V = 0 (I) [cite: 733]. Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729]. Matching 'Isochoric' with zero work done (Delta W=0) or 'Adiabatic' with zero heat exchange (Delta Q=0) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733]. Class 11 Physics: Thermodynamics

Match List-I with List-II. List-I List-II (A) Isothermal (I) Delta W (work done) =0 (B) Adiabatic (II) Delta Q (supplied heat) =0 (C) Isobaric (III) Delta U (change in internal energy) 0 (D) Isochoric

Let's evaluate each process condition based on the first law of thermodynamics: * **(A) Isothermal:** Continuous constant temperature (Delta T = 0) implies that the internal energy change of an ideal gas is zero, so Delta U = 0 (IV) [cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 (II) [cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes c

Match List-I with List-II. List-I List-II (A) Isothermal (I) Delta W (work done) =0 (B) Adiabatic (II) Delta Q…

List-I	List-II
(A) Isothermal	(I) $Delta W$ $\Delta W$ (work done) $=0$ $=0$
(B) Adiabatic	(II) $Delta Q$ $\Delta Q$ (supplied heat) $=0$ $=0$
(C) Isobaric	(III) $Delta U$ $\Delta U$ (change in internal energy) $ne0$ $\ne0$
(D) Isochoric	(IV) $Delta U=0$ $\Delta U=0$

Q5 jee_main_2025_08_april_evening Specific Heat Capacity

Water falls from a height of

200mathrm~m

into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take

g = 10mathrm~m/s^2

, specific heat of water

= 4200mathrm~J/(kgcdot K)

)

A. $0.23mathrm~K$ $0.23\mathrm{~K}$
B. $0.36mathrm~K$ $0.36\mathrm{~K}$
C. $0.14mathrm~K$ $0.14\mathrm{~K}$
D. $0.48mathrm~K$ $0.48\mathrm{~K}$

Solution

### Related Formula

Delta U = mgh quad textand quad Q = msDelta T

By conservation of energy (assuming all potential energy goes into heating the water):

mgh = msDelta T implies Delta T = fracghs

where,

g

= acceleration due to gravity

h

= height of the fall

s

= specific heat of water

Delta T

= rise in temperature ### Core Logic Given parameters: -

h = 200mathrm~m

-

g = 10mathrm~m/s^2

-

s = 4200mathrm~J/(kgcdot K)

Substitute the values to find

Delta T

:

Delta T = frac10 times 2004200 = frac20004200

Delta T = frac1021 approx 0.476mathrm~K approx 0.48mathrm~K

### Pattern Recognition Sees: "Water falling from height

h

raises temperature" → Mass cancels out.

Delta T = fracghs

. Shortcut: Always use SI units (

s_textwater = 4200mathrm~J/kgcdot K

is given; if given in

mathrmcal/gcdot^circ C

, convert using

1mathrm~cal = 4.184mathrm~J

). ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Work, Energy and Power

Q11 jee_main_2025_08_april_evening Thermodynamic Processes

A monoatomic gas having

gamma = frac53

is stored in a thermally insulated container and the gas is suddenly compressed to

left(frac18right)^mathrmth

of its initial volume. The ratio of final pressure and initial pressure is: (

gamma

is the ratio of specific heats of the gas at constant pressure and at constant volume)

A. $16$ $16$
B. $40$ $40$
C. $32$ $32$
D. $28$ $28$

Solution

### Related Formula

P_i V_i^gamma = P_f V_f^gamma

where,

P_i, P_f

= initial and final pressures

V_i, V_f

= initial and final volumes

gamma

= adiabatic exponent ### Core Logic Since the gas is stored in a "thermally insulated container" and is compressed "suddenly", the process is **adiabatic**. From the adiabatic relation:

fracP_fP_i = left(fracV_iV_fright)^gamma

Given: -

V_f = frac18 V_i implies fracV_iV_f = 8

-

gamma = frac53

### Step 1: Computation Substitute the values to find the pressure ratio:

fracP_fP_i = (8)^5/3 = left(2^3right)^5/3

fracP_fP_i = 2^5 = 32

### Pattern Recognition Sees: "suddenly compressed" or "thermally insulated container" → Adiabatic process. Shortcut:

P V^gamma = textconstant

. Since the volume goes down by

8

times, the pressure increases by

8^gamma = 8^5/3 = 32

times. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Q jee_main_2025_29_jan_evening Isothermal and Adiabatic Processes

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

**Assertion** (A): With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.

Reason (R): In isothermal process,

PV = textconstant

, while in adiabatic process

PV^gamma = textconstant

. Here

gamma

is the ratio of specific heats,

P

is the pressure and

V

is the volume of the ideal gas.

In the light of the above statements, choose the correct answer from the options given below:

A. $textBoth (A) and (R) are true but (R) is NOT the correct explanation of (A)$
B. $text(A) is true but (R) is false$ $\text{(A) is true but (R) is false}$
C. $textBoth (A) and (R) are true and (R) is the correct explanation of (A)$
D. $text(A) is false but (R) is true$ $\text{(A) is false but (R) is true}$

Solution

### Related Formula

left(fracdPdVright)_textisothermal = -fracPV

left(fracdPdVright)_textadiabatic = -gamma fracPV

### Core Logic The slope of an adiabatic process on a

P-V

diagram is

gamma

times steeper than that of an isothermal process:

left|left(fracdPdVright)_textadiabaticright| > left|left(fracdPdVright)_textisothermalright|

Isothermal and Adiabatic Processes diagram for Q2 - JEE Main 2025 Evening

When pressure increases (compression), the volume drops. Because the adiabatic curve is steeper, the pressure rises more rapidly for a given drop in volume, or conversely, for a specified increase in pressure, the volume falls off more rapidly in the isothermal process than the adiabatic process. Hence, Assertion (A) is true. Reason (R) states the governing equations

PV = C

and

PV^gamma = C

, which directly lead to these slope expressions via differentiation. Thus, Reason (R) is true and correctly explains Assertion (A). ### Pattern Recognition Adiabatic curves are steeper than isothermal curves on a

P-V

diagram because

gamma > 1

. For any expansion or compression process, remember that slope magnitude satisfies

textSlope_textadi = gamma cdot textSlope_textiso

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

Q6 jee_main_2025_29_jan_evening Heat and Work in Thermodynamic Processes

A poly-atomic molecule (

C_V = 3R, C_P = 4R

, where

R

is gas constant) goes from phase space point

A(P_A = 10^5mathrm~Pa, V_A = 4 times 10^-6mathrm~m^3)

to point

B(P_B = 5 times 10^4mathrm~Pa, V_B = 6 times 10^-6mathrm~m^3)

to point

C(P_C = 10^4mathrm~Pa, V_C = 8 times 10^-6mathrm~m^3)

.

A

to

B

is an adiabatic path and

B

to

C

is an isothermal path. The net heat absorbed per unit mole by the system is:

Heat and Work in Thermodynamic Processes diagram for Q6 - JEE Main 2025 Evening — The graph depicts a pressure vs volume plot indicating paths from state A to B (adiabatic) and from B to C (isothermal).

A. $500 mathrm~R(ln 3 + ln 4)$ $500 \mathrm{~R}(\ln 3 + \ln 4)$
B. $450 mathrm~R(ln 4 - ln 3)$ $450 \mathrm{~R}(\ln 4 - \ln 3)$
C. $500 mathrm~Rln 2$ $500 \mathrm{~R}\ln 2$
D. $400 mathrm~R ln 4$ $400 \mathrm{~R} \ln 4$

Solution

### Related Formula

Delta Q_textnet = Delta Q_AB + Delta Q_BC

Delta Q_textisothermal = nRT lnleft(fracV_fV_i ight) = P_i V_i lnleft(fracV_fV_i ight)

### Core Logic For path

A rightarrow B

: Since it is given as an adiabatic path:

Delta Q_AB = 0

For path

B rightarrow C

: Since it is given as an isothermal path, the change in internal energy

Delta U_BC = 0

. From the first law of thermodynamics, heat absorbed equals work done:

Delta Q_BC = W_BC = nRT_B lnleft(fracV_CV_B ight)

Using the ideal gas state at point

B

,

nRT_B = P_B V_B

:

P_B V_B = (5 times 10^4 mathrm~Pa) times (6 times 10^-6 mathrm~m^3) = 0.3 mathrm~J

Wait, let's express it in terms of the gas constant

R

for a single mole (

n=1

) using temperature data directly provided in the original figure labels (

T_B = 450mathrm~K

):

Delta Q_BC = (1) cdot R cdot (450) cdot lnleft(frac8 times 10^-66 times 10^-6right)

Delta Q_BC = 450 R lnleft(frac43 ight) = 450 R (ln 4 - ln 3)

Thus, the total net heat absorbed per unit mole is:

Delta Q = 0 + 450 R (ln 4 - ln 3) = 450 R (ln 4 - ln 3)

### Pattern Recognition Adiabatic paths have zero heat exchange by baseline definition. The calculation boils down directly to the work done during the isothermal stage

B rightarrow C

matching

RT ln(V_f/V_i)

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

Q7 jee_main_2025_28_jan_morning Carnot Engine and Efficiency

A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine

mathrmE_1

works between 473K to 373K and engine

mathrmE_2

works between 373K to 273K. If

eta_12

,

eta_1

and

eta_2

are the efficiencies of the engines E,

mathrmE_1

and

mathrmE_2

, respectively, then

A. $eta_12 < eta_1 + eta_2$ $\eta_{12} < \eta_1 + \eta_2$
B. $eta_12 = eta_1eta_2$ $\eta_{12} = \eta_1\eta_2$
C. $eta_12 = eta_1 + eta_2$ $\eta_{12} = \eta_{1} + \eta_{2}$
D. $eta_12 geq eta_1 + eta_2$ $\eta_{12} \geq \eta_1 + \eta_2$

Solution

### Related Formula

eta = 1 - fracmathrmT_LmathrmT_H

### Core Logic Let's compute the efficiency parameters explicitly:

eta_12 = 1 - frac273473 = frac200473 approx 0.423

eta_1 = 1 - frac373473 = frac100473 approx 0.211

eta_2 = 1 - frac273373 = frac100373 approx 0.268

Evaluating the linear sum of fractional bounds:

eta_1 + eta_2 = 0.211 + 0.268 = 0.479

Comparing the outputs clearly demonstrates:

eta_12 < eta_1 + eta_2

### Step 1: Final Conclusion Thus, the inequality satisfies option (1). ### Pattern Recognition The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as

(1-eta_12) = (1-eta_1)(1-eta_2)

, which algebraically forces

eta_12 = eta_1 + eta_2 - eta_1eta_2 < eta_1 + eta_2

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

Solution & Explanation

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