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

Q47 jee_main_2026_28_january_evening Work Done in Cyclic Process

A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is ____ J.

Work Done in Cyclic Process diagram for Q47 - JEE Main 2026 Evening — A Pressure-Volume indicator diagram showing a clockwise triangular cycle ABC.

Numerical Answer. Answer: 300 to 300

Solution

Related Formula

Wcycle = Area enclosed by the P-V cycle

Core Logic

In a P-V diagram, the work done in a cyclic process is equal to the area of the enclosed loop. Since the cycle ABC is clockwise (A $arrow$ $\rightarrow$ B $arrow$ $\rightarrow$ C $arrow$ $\rightarrow$ A), the net work done by the system is positive.

Step 1: Calculate the Area

The shape ABC is a right-angled triangle. Base of the triangle $= Δ V = VC - VA = 5 - 2 = 3 m³$ Height of the triangle $= Δ P = PB - PC = 300 - 100 = 200 Pa$

Step 2: Compute Work Done

W = (1)/(2) \times Base \times Height

W = (1)/(2) \times 3 m³ \times 200 Pa

W = 3 \times 100 = 300 J

Pattern Recognition

Always verify axis units (kPa, atm, Liters, cc) before blindly multiplying area. Here units are cleanly in standard SI (Pa and $m³$ $m^3$), so $1 Pa \cdot 1 m³ = 1 Joule$ .

Chapter Mix

Class 11 Physics: Thermodynamics

Q10 jee_main_2025_02_april_evening Adiabatic Process

Identify the characteristics of an adiabatic process in a monoatomic gas. (A) Internal energy is constant. (B) Work done in the process is equal to the change in internal energy. (C) The product of temperature and volume is a constant. (D) The product of pressure and volume is a constant. (E) The work done to change the temperature from

T₁

to

T₂

is proportional to

(T₂ - T₁)

Choose the correct answer from the options given below:

A. $(A), (C), (D) only$ $\text{(A), (C), (D) only}$
B. $(A), (C), (E) only$ $\text{(A), (C), (E) only}$
C. $(B), (E) only$ $\text{(B), (E) only}$
D. $(B), (D) only$ $\text{(B), (D) only}$

Solution

Related Formula

First Law of Thermodynamics:

$dQ = dU + dW$ $dQ = dU + dW$

In an adiabatic process:

dQ = 0 dW = -dU

Change in Internal Energy:

dU = n Cv dT = n Cv (T₂ - T₁)

Core Logic

Let's analyze each statement:

(A) Internal energy is constant: Incorrect. Since temperature changes during an adiabatic expansion/compression, internal energy ( $U \propto T$ $U \propto T$) must change.
(B) Work done is equal to the change in internal energy: Correct in magnitude ( $|dW| = |dU|$ $|dW| = |dU|$). By definition, $dW = -dU$ $dW = -dU$, which correlates the magnitude of work to the change in internal energy.
(C) Product of temperature and volume is constant: Incorrect. The adiabatic equation of state is $T Vγ-1 = constant$ $T V^{\gamma-1} = \text{constant}$.
(D) Product of pressure and volume is constant: Incorrect. The relation is $P V^γ = constant$ $P V^\gamma = \text{constant}$.
(E) Work done is proportional to $(T₂ - T₁)$ $(T_2 - T_1)$: Correct. Since $dW = -dU = -n Cv (T₂ - T₁)$ $dW = -dU = -n C_v (T_2 - T_1)$, work done is directly proportional to the temperature change $(T₂ - T₁)$ $(T_2 - T_1)$.

Step 1: Determine the correct option

Since only statements (B) and (E) are correct, the correct option is (3).

Pattern Recognition

Sees: Characteristics of adiabatic thermodynamic process. Trap: Confusing adiabatic state relations ( $PV^γ = C$ $PV^\gamma = C$, $TVγ-1 = C$ $TV^{\gamma-1} = C$) with isothermal state relations ( $PV = C$ $PV = C$, $T = C$ $T = C$). Shortcut: First law of thermodynamics under $dQ=0$ $dQ=0$ strictly enforces $|dW| = |dU|$ $|dW| = |dU|$, which validates statement B and E immediately.

Chapter Mix

Class 11 Physics: Thermodynamics

Q12 jee_main_2025_02_april_morning Thermodynamic Processes

In an adiabatic process, which of the following statements is true?

A. The molar heat capacity is infinite
B. Work done by the gas equals the increase in internal energy
C. The molar heat capacity is zero
D. The internal energy of the gas decreases as the temperature increases

Solution

Related Formula

$dQ = n C dT$ $dQ = n C dT$

dQ = 0 (for adiabatic process)

Core Logic

An adiabatic process involves no heat exchange between the system and its surroundings ( $dQ = 0$ $dQ = 0$).

The molar heat capacity $C$ $C$ is defined as:

C = (1)/(n)(dQ)/(dT)

Since $dQ = 0$ $dQ = 0$ while the temperature changes ( $dT \neq 0$ $dT \neq 0$):

$C = 0$ $C = 0$

Thus, the molar heat capacity for any adiabatic process is always zero.

Step 1: Check other options

Option (1): Isothermal processes have infinite molar heat capacity ( $dT = 0$ $dT = 0$).
Option (2): From the First Law ( $dQ = dU + dW dW = -dU$ $dQ = dU + dW \implies dW = -dU$), the work done equals the decrease in internal energy.
Option (4): The internal energy of an ideal gas ( $dU = n Cv dT$ $dU = n C_{\mathrm{v}} dT$) increases directly as temperature increases.

Step 2: Final Conclusion

The statement 'The molar heat capacity is zero' is true.

Pattern Recognition

Adiabatic = no heat flow ( $dQ=0$ $dQ=0$). Since molar heat capacity tracks the ratio of heat input to temperature change, $C$ $C$ must be 0. Conversely, isothermal has infinite capacity because heat is absorbed without any temperature change ( $dT=0$ $dT=0$).

Chapter Mix

Class 11 Physics: Thermodynamics

Q jee_main_2025_03_april_evening Thermodynamic Processes and First Law

An ideal gas exists in a state with pressure

P₀

, volume

V₀

. It is isothermally expanded to 4 times of its initial volume

(V₀)

, then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is :

A. $P₀V₀(2ln 2-0.75)$ $P_{0}V_{0}(2\ln 2-0.75)$
B. $P₀V₀(ln 2-0.75)$ $P_{0}V_{0}(\ln 2-0.75)$
C. $P₀V₀(ln 2-0.25)$ $P_{0}V_{0}(\ln 2-0.25)$
D. $P₀V₀(2ln 2-0.25)$ $P_{0}V_{0}(2\ln 2-0.25)$

Solution

Related Formula

For a cyclic thermodynamic process, the net change in internal energy is zero:

Δ Ucyclic = 0

By the First Law of Thermodynamics, the total heat exchanged $QT$ $Q_T$ equals the net work done $Wₙₑₜ$ $W_{\text{net}}$:

QT = Wₙₑₜ = W₁ + W₂ + W₃

Core Logic

The cycle consists of three steps:

Isothermal expansion from $(P₀, V₀)$ $(P_0, V_0)$ to volume $4V₀$ $4V_0$.
Isobaric compression to the original volume $V₀$ $V_0$.
Isochoric heating back to the initial state.

Step 1: Work in Isothermal Expansion ( $W₁$ $W_1$)

Initial state: $(P₀, V₀)$ $(P_0, V_0)$. Final state volume: $4V₀$ $4V_0$.

W₁ = P₀ V₀ ln((4V₀)/(V₀)) = P₀ V₀ ln(4) = 2 P₀ V₀ ln(2)

Also, the pressure at the end of this process is:

P₁ = (P₀ V₀)/(4V₀) = (P₀)/(4)

Step 2: Work in Isobaric Compression ( $W₂$ $W_2$)

The process occurs at constant pressure $P = P₁ = P₀/4$ $P = P_1 = P_0/4$. The volume goes from $4V₀$ $4V_0$ back to $V₀$ $V_0$:

W₂ = P Δ V = (P₀)/(4) (V₀ - 4V₀) = (P₀)/(4) (-3V₀) = -0.75 P₀ V₀

Step 3: Work in Isochoric Heating ( $W₃$ $W_3$)

Since the volume is held constant at $V₀$ $V_0$, no boundary work is done:

$W₃ = 0$ $W_3 = 0$

Step 4: Total Heat Exchanged ( $QT$ $Q_T$)

QT = Wₙₑₜ = W₁ + W₂ + W₃

QT = 2 P₀ V₀ ln(2) - 0.75 P₀ V₀ = P₀ V₀ (2ln(2) - 0.75)

Pattern Recognition

In any cyclic system returning to its initial state, finding total heat is mathematically equivalent to calculating the enclosed area on a P-V diagram. Here, the isobaric step occurs at the lowest expanded pressure, resulting in a simple negative rectangular area correction subtracted from the logarithmic isothermal expansion curve.

Chapter Mix

Class 11 Physics: Thermodynamics

Q jee_main_2025_07_april_morning Work Done in Thermodynamic Processes

An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is

\times 10⁻¹

J. (Take

π = 3.14

)

Elliptical thermodynamic cycle on PV plane for Q25 - JEE Main 2025 Morning — A cycle plot in which the volume lies in cm^3 (150 to 350) and pressure is in kPa (300 to 500), forming an ellipse.

Numerical Answer. Answer: 314 to 314

Solution

Related Formula

The work done $W$ $W$ in a cyclic thermodynamic process is equal to the area enclosed by the loop on a Pressure-Volume ( $P-V$ $P-V$) diagram:

W = Area of Closed Loop

For an ellipse with semi-major axis $a$ $a$ and semi-minor axis $b$ $b$:

Area = π a b

Core Logic

Determine the semi-axes of the elliptical cycle on the $P-V$ $P-V$ plane:

On the Pressure axis (x-axis):

a = Pmax - Pmin2 = (500 - 300)/(2) ~kPa = 100 ~kPa = 10⁵ ~Pa

On the Volume axis (y-axis):

b = Vmax - Vmin2 = (350 - 150)/(2) ~cm³ = 100 ~cm³ = 100 \times 10⁻⁶ ~m³ = 10⁻⁴ ~m³

Step 1: Calculate Area

Substitute $a$ $a$ and $b$ $b$ in standard SI units into the area equation:

W = π a b = 3.14 \times (10⁵ ~Pa) \times (10⁻⁴ ~m³)

W = 3.14 \times 10 = 31.4 ~J

Express in terms of $\times 10⁻¹ ~J$ $\times 10^{-1} \mathrm{~J}$:

W = 314 \times 10⁻¹ ~J

Thus, the multiplier is $314$ $314$.

Pattern Recognition

Sees: Circular/elliptical thermodynamic cycle. Shortcut: Work done is always $π Δ P Δ V / 4$ $\pi \Delta P \Delta V / 4$. Simply compute the semi-axes difference $100 ~kPa$ $100 \mathrm{~kPa}$ and $100 ~cm³$ $100 \mathrm{~cm^3}$ and multiply by $π$ $\pi$ directly, keeping tracking of metric prefixes ( $10³ \times 10⁻⁶ = 10⁻³$ $10^3 \times 10^{-6} = 10^{-3}$).

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 3

Solution

Related Formula

Core Logic

Step 1: Calculate the Area

Step 2: Compute Work Done

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine the correct option

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Check other options

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Work in Isothermal Expansion (W₁$W_1$)

Step 2: Work in Isobaric Compression (W₂$W_2$)

Step 3: Work in Isochoric Heating (W₃$W_3$)

Step 4: Total Heat Exchanged (QT$Q_T$)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Area

Pattern Recognition

Chapter Mix

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Step 1: Work in Isothermal Expansion ( $W₁$ $W_1$)

Step 2: Work in Isobaric Compression ( $W₂$ $W_2$)

Step 3: Work in Isochoric Heating ( $W₃$ $W_3$)

Step 4: Total Heat Exchanged ( $QT$ $Q_T$)