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

Q49 jee_main_2026_21_jan_morning Internal Energy

10 mole of oxygen is heated at constant volume from

30°C

to

40°C

. The change in the internal energy of the gas is ____ cal. (The molecular specific heat of oxygen at constant pressure,

Cₚ = 7 cal./mol °C

and

R = 2 cal./mol °C

.)

Numerical Answer. Answer: 500 to 500

Solution

Related Formula

Δ U = n Cv Δ T

$Cv = Cₚ - R$ $C_v = C_p - R$

Core Logic

Given values: $n = 10 moles$ $n = 10\text{ moles}$ $Δ T = 40°C - 30°C = 10°C$ $Cₚ = 7 cal/mol\cdot°C$ $R = 2 cal/mol\cdot°C$

First, find $Cv$ $C_v$ using Mayer's relation:

Cv = Cₚ - R = 7 - 2 = 5 cal/mol\cdot°C

Step 1: Calculate Internal Energy Change

Δ U = n Cv Δ T

Δ U = 10 \times (7 - 2) \times (40 - 30)

Δ U = 10 \times 5 \times 10 = 500 cal

Pattern Recognition

Change in internal energy of an ideal gas is ALWAYS $Δ U = n Cv Δ T$ $\Delta U = n C_v \Delta T$, regardless of the process (constant volume or not). Use $Cv = Cₚ - R$ $C_v = C_p - R$ when $Cₚ$ $C_p$ is given.

Chapter Mix

Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases

Q50 jee_main_2026_21_jan_evening Isobaric Process

A diatomic gas (

γ = 1.4

) does

100 J

of work when it is expanded isobarically. Then the heat given to the gas ________ J.

Numerical Answer. Answer: 350 to 350

Solution

Related Formula

W = PΔ V = nRΔ T

Q = nCₚΔ T

Cₚ = ((f)/(2) + 1)R

Core Logic

For an isobaric (constant pressure) process, work done is given by:

W = nRΔ T = 100 J

For a diatomic gas, the degrees of freedom $f = 5$ $f = 5$. Thus, the molar heat capacity at constant pressure is:

Cₚ = ((5)/(2) + 1)R = (7)/(2)R

Step 1: Calculating Heat Transfer

The heat supplied to the gas is:

Q = nCₚΔ T = n ((7)/(2)R)Δ T = (7)/(2) (nRΔ T)

Step 2: Final Conclusion

Substitute the value of work done:

Q = (7)/(2) \times (100) = 350 J

Pattern Recognition

In an isobaric process, the ratio of Work : Internal Energy Change : Heat Added ( $W : Δ U : Q$ $W : \Delta U : Q$) is always $2 : f : (f+2)$ $2 : f : (f+2)$. For diatomic gases, $f=5$ $f=5$, so the ratio is $2:5:7$ $2:5:7$. Thus $Q = (7)/(2)W$ $Q = \frac{7}{2}W$.

Chapter Mix

Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases

Q41 jee_main_2026_22_january_morning Adiabatic Process and Atomicity

The volume of an ideal gas increases 8 times and temperature becomes

(1/4)th

of initial temperature during a reversible change. If there is no exchange of heat in this process (

Δ Q = 0

) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

A. $CO₂$ $\mathrm{CO}_{2}$
B. $O₂$ $\mathrm{O}_{2}$
C. $NH₃$ $\mathrm{NH}_{3}$
D. He

Solution

Related Formula

TVγ-1 = constant

Core Logic

Solution adiabatic process diagram for Q41 - JEE Main 2026 Morning

Using adiabatic relation $TVγ-1 = constant$ $TV^{\gamma-1} = \text{constant}$:

T₁ V₁γ-1 = T₂ V₂γ-1

T(V)γ-1 = ((T)/(4))(8V)γ-1

4 = 8γ-1 2² = 23(γ-1) γ = (5)/(3)

Since $γ = 5/3$ $\gamma = 5/3$, the gas is monoatomic (Helium / He).

Pattern Recognition

Sees: Adiabatic expansion with volume and temperature changes. Shortcut: Apply $TVγ-1$ $TV^{\gamma-1}$ relation to determine adiabatic exponent $γ$ $\gamma$. Check: Matches option (4). ✓

Chapter Mix

Class 11 Physics: Thermodynamics

Q50 jee_main_2026_22_january_evening Ideal Gas Law and Gas Compression

An insulated cylinder of volume

60 ~cm³

is filled with a gas at

27^

and 2 atmospheric pressure. Then the gas is compressed making the final volume as

20 ~cm³

while allowing the temperature to rise to

77^

. The final pressure is ____ atmospheric pressure.

Numerical Answer. Answer: 7 to 7

Solution

Related Formula

(P₁ V₁)/(T₁) = (P₂ V₂)/(T₂)

T(K) = T(^ ) + 273

Core Logic

Converting initial and final temperatures to Kelvin:

T₁ = 27^ + 273 = 300 ~K

T₂ = 77^ + 273 = 350 ~K

Applying ideal gas law relation:

(2 \times 60)/(300) = (P₂ \times 20)/(350)

(120)/(300) = (20 P₂)/(350)

(2)/(5) = (20 P₂)/(350) 20 P₂ = 140 P₂ = 7 ~atm

Step 1: Final Conclusion

The final pressure is $7$ $7$ atmospheric pressure.

Pattern Recognition

Combined Gas Law: $P₂ = P₁ ((V₁)/(V₂)) ((T₂)/(T₁))$ . $P₂ = 2 \times ((60)/(20)) \times ((350)/(300)) = 2 \times 3 \times (7)/(6) = 7 atm$ .

Chapter Mix

Class 11 Physics: Thermodynamics

Q26 jee_main_2026_23_january_evening First Law of Thermodynamics

The internal energy of a monoatomic gas is

3nRT

. One mole of helium is kept in a cylinder having internal cross section area of

17 cm²

and fitted with a light movable frictionless piston. The gas is heated slowly by suppling

126 J

heat. If the temperature rises by

4°C

, then the piston will move ____ cm. (atmospheric pressure =

10⁵ Pa

)

A. $14.5$ $14.5$
B. $1.55$ $1.55$
C. $15.5$ $15.5$
D. $1.45$ $1.45$

Solution

Related Formula

Δ Q = Δ U + W

Δ U = n Cv Δ T

W = P Δ V = P (A \cdot Δ x)

Core Logic

For a monoatomic gas like Helium, $Cv = (3)/(2)R$ $C_v = \frac{3}{2}R$. Given internal energy is $3nRT$ $3nRT$, but standard change is $Δ U = n Cv Δ T$ $\Delta U = n C_v \Delta T$. Wait, the problem states internal energy $U = 3nRT$ $U = 3nRT$ (which implies $Cv = 3R$ $C_v = 3R$ for this specific state/setup, or it's a typo in the paper and they mean $U = (3)/(2)nRT$ $U = \frac{3}{2}nRT$). Let's follow the solution's lead: $Δ U = 3nRΔ T$ $\Delta U = 3nR\Delta T$ is directly given.

Step 1: Calculate Change in Internal Energy

Using the given relation:

Δ U = 3nR Δ T

Substitute $n=1$ $n=1$, $R = (25)/(3) J/mol \cdot K$ , and $Δ T = 4°C$ $\Delta T = 4^{\circ}\mathrm{C}$:

Δ U = 3 \times 1 \times (25)/(3) \times 4 = 100 J

Step 2: Apply First Law of Thermodynamics

We know $Δ Q = 126 J$ $\Delta Q = 126 \, \mathrm{J}$.

W = Δ Q - Δ U

W = 126 - 100 = 26 J

Step 3: Calculate Piston Displacement

Work done is isobaric since the piston is freely moving:

W = P Δ V = P \cdot A \cdot Δ x

26 = 10⁵ \times 17 \times 10⁻⁴ \times Δ x

26 = 170 \cdot Δ x

Δ x = (26)/(170) m = (26)/(170) \times 100 cm = 15.3 cm

Closest given option is $15.5 cm$ $15.5 \, \mathrm{cm}$.

Pattern Recognition

When dealing with a free movable piston, pressure is constant. Apply First Law directly. Watch out for modified $U$ $U$ definitions given in the prompt, replacing standard $Cv$ $C_v$ logic.

Chapter Mix

Class 11 Physics: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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Step 2: Final Conclusion

Pattern Recognition

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Solution

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Step 2: Apply First Law of Thermodynamics

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Pattern Recognition

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