Work done ($W$) in an isothermal reversible expansion of an ideal gas is given by:
\[W = -nRT \ln \left( \frac{V_2}{V_1} \right)\]
Given:
\[n = 1 \, \text{mol}, \quad T = 18^\circ \text{C} = 18 + 273.15 = 291.15 \, \text{K}\]
\[V_1 = 10 \, \text{L}, \quad V_2 = 100 \, \text{L}\]
Substitute the values:
\[W = -1 \times 0.08206 \times 291.15 \times \ln \left( \frac{100}{10} \right)\]
\[W \approx -1 \times 0.08206 \times 291.15 \times \ln(10)\]
\[W \approx -55.0128 \, \text{L atm}\]
The work done by the system is approximately $-55 \, \text{L atm}$ (rounded to nearest integer).
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