For an isobaric process, the work done is:
\[w = P \Delta V = nR \Delta T.\]
Given:
\[w = 100 \, \text{J}.\]
The heat (\(Q\)) supplied is:
\[Q = \Delta U + w,\]
where the internal energy change (\(\Delta U\)) for a diatomic gas is:
\[\Delta U = \frac{f}{2} nR \Delta T,\]
with \(f = 5\) (degrees of freedom for a diatomic gas).
Substitute:
\[Q = \frac{f}{2} nR \Delta T + nR \Delta T.\]
Simplify:
\[Q = \left(\frac{f}{2} + 1\right) nR \Delta T.\]
Substitute \(\Delta T\) from \(w = nR \Delta T\):
\[Q = \left(\frac{f}{2} + 1\right) \times w.\]
Substitute \(f = 5\) and \(w = 100 \, \text{J}\):
\[Q = \left(\frac{5}{2} + 1\right) \times 100 = \left(\frac{7}{2}\right) \times 100 = 350 \, \text{J}.\]
Thus, the heat given to the gas is:
\[Q = 350 \, \text{J}.\]
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: