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Thermodynamics appeared 36 times across 3 years — 4.2% of Physics. This question is from Work Done in Thermodynamic Processes.

Year 2026 2025 2024 Total
Questions 11 19 6 36

An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is × 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 Type:
Enter a numerical value Answer: 314 to 314 +4 marks

Solution & Explanation

Related Formula

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

W = Area of Closed Loop

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

Area = π a b
Core Logic

Determine the semi-axes of the elliptical cycle on the 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 × 10⁻⁶ ~m³ = 10⁻⁴ ~m³
Step 1: Calculate Area

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

W = π a b = 3.14 × (10⁵ ~Pa) × (10⁻⁴ ~m³) W = 3.14 × 10 = 31.4 ~J

Express in terms of × 10⁻¹ ~J:

W = 314 × 10⁻¹ ~J

Thus, the multiplier is 314.

Pattern Recognition

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

Chapter Mix

Class 11 Physics: Thermodynamics

PV integration and area calculation details
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.

More Thermodynamics Previous-Year Questions — Page 8

Q jee_main_2024_31_jan_morning Isobaric Process
The given figure represents two isobaric processes for the same mass of an ideal gas, then
Isobaric Process diagram for Q38 - JEE Main 2024 Morning
A Volume vs Temperature (V-T) graph showing two straight lines starting from the origin representing distinct constant pressures P1 and P2.
  • A. P₂≥ P₁
  • B. P₂ > P₁
  • C. P₁ = P₂
  • D. P₁ > P₂

Solution

Related Formula

PV = nRT

Core Logic

From the Ideal Gas Law:

V = ((nR)/(P)) T

In a V-T graph, the equation of the line represents y = mx, where the slope m is:

Slope = (nR)/(P) Slope ∝ (1)/(P)

Thus, a higher slope corresponds to a lower pressure.

Step 2: Compare Slopes

From the given figure, the slope of line 2 is greater than the slope of line 1:

(Slope)₂ > (Slope)₁

Therefore, inversely: P₂ < P₁ or P₁ > P₂.

Pattern Recognition

In V-T graphs, steeper lines mean lower Pressure. In P-T graphs, steeper lines mean lower Volume. It's an inverse inverse slope relationship.

Chapter Mix

Class 11 Physics: Thermodynamics

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