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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 :-

Solution & Explanation

### Related Formula From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: Delta Q = Delta U + W ### Core Logic The question states that pressure increases linearly with temperature, which means their ratio is constant : P = kT implies fracPT = textconstant Since fracPT = fracnRV, the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). ### Step 1: Evaluate All Statements * Statement A: True. In an isochoric process, dV = 0 implies W = int 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. ### Step 2: Final Selection Gathering the true statements (A, D, and E) points directly to Option (2). ### Pattern Recognition 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Reference Study Guides

Thermodynamics

More Thermodynamics Previous-Year Questions — Page 4

Q12 2025 Adiabatic Compression
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800~mathrmcm^3 and temperature 27^circmathrmC . The change in temperature when the gas is adiabatically compressed to 200~mathrmcm^3 is: (Take gamma = 1.5)
  • A. 327mathrm~K
  • B. 600mathrm~K
  • C. 522mathrm~K
  • D. 300mathrm~K

Solution

### Related Formula For an adiabatic process: T V^gamma - 1 = textconstant where, T = absolute temperature in Kelvin, V = volume of the gas, gamma = adiabatic exponent. ### Core Logic Given values: - Initial volume, V_1 = 800mathrm~cm^3 - Final volume, V_2 = 200mathrm~cm^3 - Initial temperature, T_1 = 27^circmathrmC = 27 + 273 = 300mathrm~K - Adiabatic exponent, gamma = 1.5 implies gamma - 1 = 0.5 ### Step 1: Calculating Final Temperature Apply the adiabatic relation: T_1 V_1^gamma - 1 = T_2 V_2^gamma - 1 T_2 = T_1 left(fracV_1V_2right)^gamma - 1 Substitute the values: T_2 = 300 left(frac800200right)^0.5 = 300 times (4)^0.5 T_2 = 300 times 2 = 600mathrm~K ### Step 2: Calculating Change in Temperature Now compute the change in temperature (Delta T): Delta T = T_2 - T_1 = 600mathrm~K - 300mathrm~K = 300mathrm~K ### Pattern Recognition Always read carefully to see if the question asks for the **final temperature** or the **change in temperature**. Many students lose marks by choosing 600mathrm~K (the final temperature) instead of the difference 300mathrm~K! Stay sharp. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q17 2025 Thermodynamic Processes
Match List-I with List-II:
List-IList-II
(A) Isobaric(I) Delta Q=Delta W
(B) Isochoric(II) Delta Q=Delta U
(C) Adiabatic(III) Delta Q=textzero
(D) Isothermal(IV) Delta Q=Delta U+PDelta V
Choose the correct answer from the options given below:
  • A. \text{(A)-(IV), (B)-(III), (C)-(II), (D)-(I)}
  • B. \text{(A)-(IV), (B)-(I), (C)-(III), (D)-(II)}
  • C. \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
  • D. \text{(A)-(II), (B)-(IV), (C)-(III), (D)-(I)}

Solution

### Related Formula First Law of Thermodynamics: Delta Q = Delta U + Delta W ### Core Logic - **Isobaric**: Pressure is constant, work done Delta W = PDelta V. Thus, Delta Q = Delta U + PDelta V (Matches IV). - **Isochoric**: Volume is constant, Delta V = 0 implies Delta W = 0. Thus, Delta Q = Delta U (Matches II). - **Adiabatic**: No heat transfer, Delta Q = 0 (Matches III). - **Isothermal**: Temperature is constant, internal energy change Delta U = 0 for an ideal gas. Thus, Delta Q = Delta W (Matches I). ### Step 1: Match Compilation Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I). ### Pattern Recognition Adiabatic definition is always heat-isolated (Q=0), isochoric implies rigid boundaries (W=0). Identifying these two immediately isolates the correct match permutation in seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q17 2025 Thermodynamic Processes
Match List-I with List-II.
List-IList-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) ne0
(D) Isochoric(IV) Delta U=0
Choose the correct answer from the options given below: [cite: 151, 152]
  • A. (A)-(III), (B)-(II), (C)-(I), (D)-(IV) [cite: 153]
  • B. (A)-(IV), (B)-(I), (C)-(III), (D)-(II) [cite: 154]
  • C. (A)-(IV), (B)-(II), (C)-(III), (D)-(I) [cite: 155]
  • D. (A)-(II), (B)-(IV), (C)-(I), (D)-(III) [cite: 156]

Solution

### Core Logic 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 implies text(IV) [cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 implies text(II) [cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, Delta U neq 0 implies text(III) [cite: 732]. * **(D) Isochoric:** Rigid boundary condition at constant volume (Delta V = 0) ensures work done Delta W = PDelta V = 0 implies text(I) [cite: 733]. Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729]. ### Pattern Recognition 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]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q6 2025 Adiabatic Processes
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases. Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process. In the light of the above statement, choose the correct answer from the options given below :
  • A. Both (A) and (R) are true and (R) is the correct explanation of (A)
  • B. (A) is true but (R) is false
  • C. (A) is false but (R) is true
  • D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

Solution

### Related Formula T_1 V_1^gamma-1 = T_2 V_2^gamma-1 ### Core Logic Assertion (A) review: An insulated container means the process is adiabatic (Q=0). When the gas is shrunk (compressed) to half its volume (V_2 = V_1 / 2), T_2 = T_1 left(fracV_1V_2 ight)^gamma-1 = T_1 (2)^gamma-1 Since gamma > 1, T_2 > T_1. Therefore, temperature increases during adiabatic compression, which makes Assertion (A) false. Reason (R) review: Free expansion occurs when a gas expands into a vacuum inside an insulated container. No work is done (W=0) and no heat is exchanged (Q=0), hence it is adiabatic. It cannot spontaneously reverse, so it is irreversible. Thus, Reason (R) is true. ### Pattern Recognition Adiabatic compression always raises temperature due to work being done on the gas, while free expansion keeps the temperature of an ideal gas constant (dI=0 as W=0, Q=0). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q13 2025 Cyclic Processes
The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure
Cyclic semi-circular P-V process indicator diagram Q13
The image shows a P-V cycle consisting of a horizontal line from C to A and a semicircular loop from A back to C via B.
) is (in SI unit)
  • A. 10pi
  • B. 5pi
  • C. zero
  • D. 40pi

Solution

### Related Formula From the first law of thermodynamics for a complete cycle: Delta U = 0 implies Q = W = textArea of the loop ### Core Logic The graph shows a semicircle in a Ptext-V indicator diagram.
P-V cycle geometry calculation graph Q13
The image shows a P-V cycle consisting of a horizontal line from C to A and a semicircular loop from A back to C via B.
- Pressure dimension diameter: Delta P = 400 - 200 = 200\ mathrmkPa = 200 times 10^3\ mathrmPa - Volume dimension diameter: Delta V = 400 - 200 = 200\ mathrmcc = 200 times 10^-6\ mathrmm^3 Radius along Pressure axis: R_P = 100 times 10^3\ mathrmPa Radius along Volume axis: R_V = 100 times 10^-6\ mathrmm^3 Area of the closed semicircular path: W = frac12 pi R_P R_V W = frac12 times pi times (100 times 10^3) times (100 times 10^-6) W = frac10pi2 = 5pi\ mathrmJ Since Q = W, the heat exchanged has a magnitude of 5pi\ mathrmJ. ### Pattern Recognition For a cycle on an indicator chart with mismatched scales, use the elliptic area template pi a b (or frac12pi a b for a half-ellipse/semicircle). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

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