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 $ \pi = 3.14 $)
The graph is a circle in the PV diagram, and for a cyclic process, the work done by the gas is equal to the area enclosed by the cycle.
Assuming both axes are scaled equally, the radius \( R = 100 \) \[ \text{Area} = \pi R^2 = 3.14 \times 100^2 = 3.14 \times 10^4 \] Convert units: \[ 1\, \text{kPa} \cdot \text{cm}^3 = 10^{-2} \, \text{J} \Rightarrow \text{Work done} = 3.14 \times 10^4 \times 10^{-2} = 314 \, \text{J} \]
Work done = $31.4 \times 10^{-1}$ J
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: