The osmotic pressure of a dilute solution is Pa at 273 K. Osmotic pressure of the same solution at 283 K is

Question

The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

Smart Achievers · Accepted Answer

(73)

$Osmotic pressure(π) (for a non - electrolyte solution) is related to concentration (C) and temperature (T) as,$

$π = C R T, where R is universal gas constant$

$π_{1} = C R T_{1}, π_{2} = C R T_{2}$

$\frac{π_{2}}{π_{1}} = \frac{T_{2}}{T_{1}}$

$\frac{π_{2}}{7 \times 10^{5} Pa} = \frac{283 K}{273 K}$

$π_{2} = 7 \times 10^{5} \times \frac{283}{273} Pa$

$= 7.256 \times 10^{5} {Nm}^{- 2}$

$= 72.56 \times 10^{4} {Nm}^{- 2}$

Solution

Q.
The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

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Solution

Q. The osmotic pressure of a dilute solution is 7×105 Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ ×104 Nm-2. [2024]

Ans. (73) Osmotic pressure(π) (for a non-electrolyte solution) is related to concentration (C) and temperature (T) as, π=CRT, where R is universal gas constant π1=CRT1, π2=CRT2 π2π1=T2T1 π27×105 Pa=283 K273 K π2=7×105×283273 Pa =7.256×105 Nm-2 =72.56×104 Nm-2

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Q.
The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]