Question

The osmotic pressure of a dilute solution is \(7 × 10^5\) Pa at \(273K\). Osmotic pressure of the same solution at\(283K\) is _____\(× 10^4 Nm^2\)

Answer 1

Correct Answer: 72.56

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

Answer 2

Step 1: Formula for osmotic pressure.

For a given solution (same concentration): \[ \pi = C R T \] Since \( C \) and \( R \) are constants for the same solution, \[ \frac{\pi_1}{T_1} = \frac{\pi_2}{T_2} \] \[ \Rightarrow \pi_2 = \pi_1 \frac{T_2}{T_1} \]

Step 2: Substitute given values.

\[ \pi_1 = 7 \times 10^5 \, \text{Pa}, \quad T_1 = 273\,\text{K}, \quad T_2 = 283\,\text{K} \] \[ \pi_2 = 7 \times 10^5 \times \frac{283}{273} \]

Step 3: Simplify the expression.

\[ \pi_2 = 7 \times 10^5 \times 1.0366 = 7.256 \times 10^5 \, \text{Pa} \] \[ \pi_2 = 72.56 \times 10^4 \, \text{N/m}^2 \]

Final Answer:

\[ \boxed{72.56 \times 10^4 \, \text{N/m}^2} \]