Skip to content

#vapor pressure

48 public questions tagged with this topic.

In a closed container, water is in equilibrium with its vapor at 25°C, and the vapor pressure is 0.0313 atm. If the vol

Given: In a closed container, water is in equilibrium with its vapor at 25°C, and the vapor pressure is 0.0313 atm. If the volume of the container is doubled suddenly, what happens to the vapor pressure initially? These values define the system as per NCERT data. Formula: Initially, the number of vapor molecules remains the same, but the volume doubles, reducing the pressure to half: 0.0313 / 2 = 0.01565 approx 0.0157 atm.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂N

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

The vapor pressure of pure water is 25 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor

(p⁰ - p/p⁰) = xsolute . (25 - 23/25) = 0.08 . Moles of water = (180/18) = 10 . xsolute = (nsolute/nsolute + 10) = 0.08 . nsolute = 0.08 (nsolute + 10) , nsolute - 0.08 nsolute = 0.8 , 0.92 nsolute = 0.8 , nsolute ≈ 0.8696 . Mass = 0.8696 × 60 ≈ 52.18 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions