The relation between heat at constant pressure (\( \Delta H \)) and at constant volume (\( \Delta U \)) is:
\( \Delta H = \Delta U + \Delta n_g RT \)
For benzoic acid:
\( C_6H_5COOH(s) + \frac{15}{2} O_2(g) \rightarrow 7CO_2(g) + 3H_2O(l) \)
\( \Delta n_g = 7 - \frac{15}{2} = -\frac{1}{2} \). Substituting:
\( \Delta H = -321.30 - \frac{1}{2} R \times 300 \)
Here, \( R \approx 8.314 \, \text{J/mol.K} \). Solving gives:
\( x = 150 \)
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