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The solubility of a gas is 0.980 at a pressure of 116 kPa. What is the solubility of the gas if the pressure is decreased to 101 kPa, given that the temperature is held constant

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Eduard22sly

The solubility of the gas when the pressure is decreased to 101 KPa at constant temperature is 0.853

Henry's law

This states that the solubility of a gas in liquids at constant temperature is proportional to its pressure. Mathematically, it can be expressed as

S₁ / P₁ = S₂ / P₂

Where

  • S₁ is the initial solubility
  • P₁ is the initial pressure
  • S₂ is the new solubility
  • P₂ is the new pressure

How to determine the new solubility

From the question given above, the following data were obtained

  • Initial solubility (S₁) = 0.980
  • Initial pressure (P₁) = 116 KPa
  • New pressure (P₂) = 101 KPa
  • New solubility (S₂) =?

S₁ / P₁ = S₂ / P₂

0.98 / 116 = S₂ / 101

Cross multiply

116 × S₂ = 0.98 × 101

Divide both side by 116

S₂ = (0.98 × 101) / 116

S₂ = 0.853

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