The osmotic pressure (\(\pi\)) is given by the formula:
\(\pi = CRT\)
Since concentration (C) and R are constants, we can use the ratio:
\(\frac{\pi_1}{\pi_2} = \frac{T_1}{T_2}\)
Rearranging to find \(\pi_2\):
\(\pi_2 = \pi_1 \cdot \frac{T_2}{T_1} = 7 \times 10^5 \times \frac{283}{273}\)
Calculating \(\pi_2\):
\(\pi_2 = 72.56 \times 10^4 \, \text{Nm}^{-2}\)
Thus, the osmotic pressure at 283 K is approximately:
So, the correct answer is: 72.56 or 73
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