Question

the osmatic pressure of sodium at 27०c is 6atm what will be the osmatic pressure at 600k under similar conditions?​

Answer 1

Given :

Osmotic pressure of sodium at 27°C = 6atm

To Find :

Osmotic pressure of sodium at 600k.

Solution :

❒ The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called osmotic pressure.

• Gay Lussac's law (For ideal solution) : At constant volume, pressure of a fixed amount of a gas varies directly with the temperature$.$
• P/P' = T/T'

We know that, 0°C = 273k

Therefore,

• T = 27 + 273 = 300k

By plugging the given values,

➠ P/P' = T/T'

➠ 6/P' = 300/600

➠ P' = 6 × 6/3

➠ P' = 12 atm

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

The osmatic pressure of sodium at 27०c is 6atm what will be the osmatic pressure at 600k under similar conditions?

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• Osmatic pressure of sodium at 27°C is 6atm.

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• Osmatic pressure if sodium at 600k.

${\bigstar}\large{\boxed{\sf{\pink{Osmosis \: pressure}}}}$

The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semiperamble is known to be osmosis pressure.

${\bigstar}\large{\boxed{\sf{\pink{Solution}}}}$

$\large\red{\texttt{Gay lussac's law}}$ $\large\green{\texttt{For ideal solution}}$ At constant volume pressure of a fixed amount of a gas varies directly with the temperature.

$\large\pink{\texttt{P/P¹ = T/T¹}}$

We know that 0°C = 273k

Therefore,

• T = 27 + 273 = 300k

By substituting the given value,

• P/P¹ = T/T¹
• 6/P¹ = 300/600
• P¹ = 6 × 6/3
• P¹ = 12atm

$\large\orange{\texttt{12 atm is the answer}}$

@Itzbeautyqueen23

Hope it's helpful.

Thank you :)