The osmotic pressure of a dilute solution is $7 \times 10^5 \mathrm{~Pa}$ at $273 \mathrm{~K}$. Osmotic pressure of the same solution at $283 \mathrm{~K}$ is _________ $\times 10^4 \mathrm{Nm}^{-2}$.
Answer (integer)
73
Solution
<p>To calculate the osmotic pressure of the solution at the new temperature, we can use the formula:</p>
<p>$ \pi = \text{CRT} $</p>
<p>Since the concentration $ C $ and the gas constant $ R $ remain constant, the ratio of the osmotic pressures at two different temperatures is given by:</p>
<p>$ \frac{\pi_1}{\pi_2} = \frac{T_1}{T_2} $</p>
<p>Thus, we can express the osmotic pressure at the second temperature as:</p>
<p>$ \pi_2 = \frac{\pi_1 \cdot T_2}{T_1} $</p>
<p>Substituting the given values:</p>
<p>$ \pi_2 = \frac{7 \times 10^5 \times 283}{273} $</p>
<p>After calculation, this simplifies to:</p>
<p>$ \pi_2 = 72.56 \times 10^4 \, \text{Nm}^{-2} $</p>
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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