A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine E1 works between 473K to 373K and engine E2 works between 373K to 273K. If 12 , 1 and 2 are the efficiencies of the engines E, E1 and E2 , respectively, then

Let's compute the efficiency parameters explicitly: Evaluating the linear sum of fractional bounds: Comparing the outputs clearly demonstrates: Thus, the inequality satisfies option (1). The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as (1-12) = (1-1)(1-2), which algebraically forces 12 = 1 + 2 - 12 < 1 + 2. Class 11 Physics: Thermodynamics

A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine E1 works between 473K to 373K and engine E2 works between 373K to 273K. If 12 , 1 and 2 are

Let's compute the efficiency parameters explicitly: Evaluating the linear sum of fractional bounds: Comparing the outputs clearly demonstrates: Thus, the inequality satisfies option (1). The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as (1-12) = (1-1)(1-2), which algebraically forces 12 = 1 + 2 - 12 < 1 + 2. Class 11 Physics: Thermodynamics

A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine E1 works between 473K to 373K and engine E2 works between 373K to 273K. If 12 , 1 and 2 are the efficiencies of the engines E, E1 and E2 , respectively, then

Let's compute the efficiency parameters explicitly: Evaluating the linear sum of fractional bounds: Comparing the outputs clearly demonstrates: Thus, the inequality satisfies option (1). The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as (1-12) = (1-1)(1-2), which algebraically forces 12 = 1 + 2 - 12 < 1 + 2. Class 11 Physics: Thermodynamics

A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine E1 works between 473K to 373K and engine E2 works between 373K to 273K. If 12 , 1 and 2 are

Let's compute the efficiency parameters explicitly: Evaluating the linear sum of fractional bounds: Comparing the outputs clearly demonstrates: Thus, the inequality satisfies option (1). The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as (1-12) = (1-1)(1-2), which algebraically forces 12 = 1 + 2 - 12 < 1 + 2. Class 11 Physics: Thermodynamics

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

Year	2026	2025	2024	Total
Questions	11	19	6	36

More Thermodynamics Previous-Year Questions — Page 4

Q5 jee_main_2025_08_april_evening Specific Heat Capacity

Water falls from a height of

200~m

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

g = 10~m/s²

, specific heat of water

= 4200~J/(kg\cdot K)

)

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

Solution

Related Formula

Δ U = mgh and Q = msΔ T

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

mgh = msΔ T Δ T = (gh)/(s)

where, $g$ $g$ = acceleration due to gravity $h$ $h$ = height of the fall $s$ $s$ = specific heat of water $Δ T$ $\Delta T$ = rise in temperature

Core Logic

Given parameters:

$h = 200~m$ $h = 200\mathrm{~m}$
$g = 10~m/s²$ $g = 10\mathrm{~m/s}^{2}$
$s = 4200~J/(kg\cdot K)$ $s = 4200\mathrm{~J/(kg\cdot K)}$

Substitute the values to find $Δ T$ $\Delta T$:

Δ T = (10 \times 200)/(4200) = (2000)/(4200)

Δ T = (10)/(21) \approx 0.476~K \approx 0.48~K

Pattern Recognition

Sees: "Water falling from height $h$ $h$ raises temperature" → Mass cancels out. $Δ T = (gh)/(s)$ $\Delta T = \frac{gh}{s}$. Shortcut: Always use SI units ( $swater = 4200~J/kg\cdot K$ is given; if given in $cal/g\cdot^° C$ $\mathrm{cal/g\cdot^\circ C}$, convert using $1~cal = 4.184~J$ $1\mathrm{~cal} = 4.184\mathrm{~J}$). ✓

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

γ = (5)/(3)

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

((1)/(8))th

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

γ

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

Pᵢ Vᵢγ = Pf Vfγ

where, $Pᵢ, Pf$ $P_i, P_f$ = initial and final pressures $Vᵢ, Vf$ $V_i, V_f$ = initial and final volumes $γ$ $\gamma$ = 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:

(Pf)/(Pᵢ) = ((Vᵢ)/(Vf))γ

Given:

$Vf = (1)/(8) Vᵢ (Vᵢ)/(Vf) = 8$
$γ = (5)/(3)$ $\gamma = \frac{5}{3}$

Step 1: Computation

Substitute the values to find the pressure ratio:

(Pf)/(Pᵢ) = (8)5/3 = (2³)5/3

(Pf)/(Pᵢ) = 2⁵ = 32

Pattern Recognition

Sees: "suddenly compressed" or "thermally insulated container" → Adiabatic process. Shortcut: $P V^γ = constant$ $P V^\gamma = \text{constant}$. Since the volume goes down by $8$ $8$ times, the pressure increases by $8γ = 85/3 = 32$ $8^{\gamma} = 8^{5/3} = 32$ times. ✓

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 = constant

, while in adiabatic process

PVγ = constant

. Here

γ

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

Solution

Related Formula

((dP)/(dV))isothermal = -(P)/(V)

((dP)/(dV))adiabatic = -γ (P)/(V)

Core Logic

The slope of an adiabatic process on a $P-V$ $P-V$ diagram is $γ$ $\gamma$ times steeper than that of an isothermal process:

|((dP)/(dV))adiabatic| > |((dP)/(dV))isothermal|

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 $PV = C$ $PV = C$ and $PVγ = C$ $PV^{\gamma} = C$, 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 $P-V$ $P-V$ diagram because $γ > 1$ $\gamma > 1$. For any expansion or compression process, remember that slope magnitude satisfies $Slopeadi = γ \cdot Slopeᵢₛₒ$ .

Chapter Mix

Class 11 Physics: Thermodynamics

Q6 jee_main_2025_29_jan_evening Heat and Work in Thermodynamic Processes

A poly-atomic molecule (

CV = 3R, CP = 4R

, where

R

is gas constant) goes from phase space point

A(PA = 10⁵~Pa, VA = 4 \times 10⁻⁶~m³)

to point

B(PB = 5 \times 10⁴~Pa, VB = 6 \times 10⁻⁶~m³)

to point

C(PC = 10⁴~Pa, VC = 8 \times 10⁻⁶~m³)

A

B

is an adiabatic path and

B

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

Solution

Related Formula

Δ Qₙₑₜ = Δ QAB + Δ QBC

Δ Qisothermal = nRT ln((Vf)/(Vᵢ)) = Pᵢ Vᵢ ln((Vf)/(Vᵢ))

Core Logic

For path $A arrow B$ $A \rightarrow B$: Since it is given as an adiabatic path:

Δ QAB = 0

For path $B arrow C$ $B \rightarrow C$: Since it is given as an isothermal path, the change in internal energy $Δ UBC = 0$ $\Delta U_{BC} = 0$. From the first law of thermodynamics, heat absorbed equals work done:

Δ QBC = WBC = nRTB ln((VC)/(VB))

Using the ideal gas state at point $B$ $B$, $nRTB = PB VB$ $nRT_B = P_B V_B$:

PB VB = (5 \times 10⁴ ~Pa) \times (6 \times 10⁻⁶ ~m³) = 0.3 ~J

Wait, let's express it in terms of the gas constant $R$ $R$ for a single mole ( $n=1$ $n=1$) using temperature data directly provided in the original figure labels ( $TB = 450~K$ $T_B = 450\mathrm{~K}$):

Δ QBC = (1) \cdot R \cdot (450) \cdot ln( 8 \times 10⁻⁶6 \times 10⁻⁶)

Δ QBC = 450 R ln((4)/(3)) = 450 R (ln 4 - ln 3)

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

Δ 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 $B arrow C$ $B \rightarrow C$ matching $RT ln(Vf/Vᵢ)$ $RT \ln(V_f/V_i)$.

Chapter Mix

Class 11 Physics: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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More Thermodynamics Previous-Year Questions — Page 4

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Related Formula

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