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$$W = \text{Area of Closed Loop}$$
For an ellipse with semi-major axis a$a$ and semi-minor axis b$b$:
Area = π a b$$\text{Area} = \pi a b$$
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$$a = \frac{P_{\text{max}} - P_{\text{min}}}{2} = \frac{500 - 300}{2} \mathrm{~kPa} = 100 \mathrm{~kPa} = 10^5 \mathrm{~Pa}$$
- On the Volume axis (y-axis):
b = Vmax - Vmin2 = (350 - 150)/(2) ~cm³ = 100 ~cm³ = 100 × 10⁻⁶ ~m³ = 10⁻⁴ ~m³$$b = \frac{V_{\text{max}} - V_{\text{min}}}{2} = \frac{350 - 150}{2} \mathrm{~cm}^3 = 100 \mathrm{~cm}^3 = 100 \times 10^{-6} \mathrm{~m}^3 = 10^{-4} \mathrm{~m}^3$$
Step 1: Calculate Area
Substitute a$a$ and b$b$ in standard SI units into the area equation:
W = π a b = 3.14 × (10⁵ ~Pa) × (10⁻⁴ ~m³)$$W = \pi a b = 3.14 \times (10^5 \mathrm{~Pa}) \times (10^{-4} \mathrm{~m}^3)$$
W = 3.14 × 10 = 31.4 ~J$$W = 3.14 \times 10 = 31.4 \mathrm{~J}$$
Express in terms of × 10⁻¹ ~J$\times 10^{-1} \mathrm{~J}$:
W = 314 × 10⁻¹ ~J$$W = 314 \times 10^{-1} \mathrm{~J}$$
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 × 10⁻⁶ = 10⁻³$10^3 \times 10^{-6} = 10^{-3}$).
Chapter Mix
Class 11 Physics: Thermodynamics