An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process) Choose the correct answer from the options given below :-

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning heat added equals the change in internal energy. * Statement C: False. Volume is constant, so it does not increase. * Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase. Gathering the true statements (A, D, and E) points directly to Option (2). A linear P-T line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation. Class 11 Physics: Thermodynamics

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat add

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning he

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process) Choose the correct answer from the options given below :-

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning heat added equals the change in internal energy. * Statement C: False. Volume is constant, so it does not increase. * Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase. Gathering the true statements (A, D, and E) points directly to Option (2). A linear P-T line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation. Class 11 Physics: Thermodynamics

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat add

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning he

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with…

Year	2026	2025	2024	Total
Questions	11	19	6	36

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

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. $500 ~R(ln 3 + ln 4)$ $500 \mathrm{~R}(\ln 3 + \ln 4)$
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

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

E₁

works between 473K to 373K and engine

E₂

works between 373K to 273K. If

η₁₂

,

η₁

and

η₂

are the efficiencies of the engines E,

E₁

and

E₂

, respectively, then

A. $η₁₂ < η₁ + η₂$ $\eta_{12} < \eta_1 + \eta_2$
B. $η₁₂ = η₁η₂$ $\eta_{12} = \eta_1\eta_2$
C. $η₁₂ = η₁ + η₂$ $\eta_{12} = \eta_{1} + \eta_{2}$
D. $η₁₂ \geq η₁ + η₂$ $\eta_{12} \geq \eta_1 + \eta_2$

Solution

Related Formula

η = 1 - TLTH

Core Logic

Let's compute the efficiency parameters explicitly:

η₁₂ = 1 - (273)/(473) = (200)/(473) \approx 0.423

η₁ = 1 - (373)/(473) = (100)/(473) \approx 0.211

η₂ = 1 - (273)/(373) = (100)/(373) \approx 0.268

Evaluating the linear sum of fractional bounds:

η₁ + η₂ = 0.211 + 0.268 = 0.479

Comparing the outputs clearly demonstrates:

η₁₂ < η₁ + η₂

Step 1: Final Conclusion

Thus, the inequality satisfies option (1).

Pattern Recognition

The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as $(1-η₁₂) = (1-η₁)(1-η₂)$ $(1-\eta_{12}) = (1-\eta_1)(1-\eta_2)$, which algebraically forces $η₁₂ = η₁ + η₂ - η₁η₂ < η₁ + η₂$ .

Chapter Mix

Class 11 Physics: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate All Statements

Step 2: Final Selection

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 4

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