For an adiabatic process, the relation between pressure and volume is:
\[ P_iV_i^\gamma = P_fV_f^\gamma, \] where \(\gamma = 1.5\).
Substitute the given volumes (\(V_i = 5 \, \text{litres}, V_f = 4 \, \text{litres}\)):
\[ P_i (5)^{1.5} = P_f (4)^{1.5}. \]
Rearranging for \(\frac{P_i}{P_f}\):
\[ \frac{P_i}{P_f} = \frac{(4)^{1.5}}{(5)^{1.5}}. \]
Simplify:
\[ \frac{P_i}{P_f} = \left(\frac{4}{5}\right)^{1.5}. \]
Write \(1.5\) as \(\frac{3}{2}\):
\[ \frac{P_i}{P_f} = \left(\frac{4}{5}\right)^{\frac{3}{2}} = \left(\frac{4}{5}\right)^1 \times \left(\frac{4}{5}\right)^{\frac{1}{2}}. \]
Simplify each term:
\[ \frac{P_i}{P_f} = \frac{4}{5} \times \sqrt{\frac{4}{5}} = \frac{4}{5} \times \frac{2}{\sqrt{5}} = \frac{8}{5\sqrt{5}}. \]
Final Answer: \(\frac{8}{5\sqrt{5}}\).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32