JEE Challenger
More from States of Matter
ChemMedAdv 2011 • Paper 1

Calculate Total Volume of Evacuated Vessel with Helium and Unknown Compound

To an evacuated vessel with movable piston under external pressure of 1 atm.1\text{ atm.}, 0.1 mol0.1\text{ mol} of He\text{He} and 1.0 mol1.0\text{ mol} of an unknown compound (vapour pressure 0.68 atm.0.68\text{ atm.} at 0∘C0^\circ\text{C}) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0∘C0^\circ\text{C} is close to

Official Numerical Answer7

Topics & Concepts

States of MatterIdeal Gas LawReal Gases

Step-by-Step Solution

To find the total volume of the gases at 0∘C0^\circ\text{C}, we apply the ideal gas law and Dalton's Law of Partial Pressures.

1. Equilibrium Conditions

Since the vessel has a movable piston under a constant external pressure of 1 atm1\text{ atm}, the total pressure of the gas mixture inside the vessel at equilibrium must equal the external pressure:
Ptotal=1.0 atmP_{\text{total}} = 1.0\text{ atm}

2. Partial Pressures of the Components

If liquid and vapour co-exist for the unknown compound, its partial pressure in the gas phase equals its saturation vapour pressure at 0∘C0^\circ\text{C}:
Pcompound=0.68 atmP_{\text{compound}} = 0.68\text{ atm}

By Dalton's Law of Partial Pressures:
Ptotal=PHe+PcompoundP_{\text{total}} = P_{\text{He}} + P_{\text{compound}}

Substitute the known values to find the partial pressure of Helium (PHeP_{\text{He}}):
PHe=1.0 atm−0.68 atm=0.32 atmP_{\text{He}} = 1.0\text{ atm} - 0.68\text{ atm} = 0.32\text{ atm}

3. Calculation of Total Volume

Since Helium is a non-condensable gas, it occupies the entire volume VV of the vessel. Using the ideal gas equation for Helium:
PHe⋅V=nHeRTP_{\text{He}} \cdot V = n_{\text{He}} R T

Given:

  • nHe=0.1 moln_{\text{He}} = 0.1\text{ mol}
  • T=0∘C=273.15 KT = 0^\circ\text{C} = 273.15\text{ K}
  • RT≈22.414 L⋅atm⋅mol−1R T \approx 22.414\text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}

Substitute these values into the equation:
V=nHeRTPHe=0.1 mol×22.414 L⋅atm⋅mol−10.32 atmV = \frac{n_{\text{He}} R T}{P_{\text{He}}} = \frac{0.1\text{ mol} \times 22.414\text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}}{0.32\text{ atm}}

V=2.24140.32≈7.004 LV = \frac{2.2414}{0.32} \approx 7.004\text{ L}

4. Verification of Liquid-Vapour Equilibrium

To ensure that condensation actually occurs, we calculate the number of moles of the unknown compound present in the vapour phase (nvaporn_{\text{vapor}}):
nvapor=PcompoundVRT=0.68 atm×7.004 L22.414 L⋅atm⋅mol−1=0.2125 moln_{\text{vapor}} = \frac{P_{\text{compound}} V}{R T} = \frac{0.68\text{ atm} \times 7.004\text{ L}}{22.414\text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}} = 0.2125\text{ mol}

Since nvapor=0.2125 mol<1.0 moln_{\text{vapor}} = 0.2125\text{ mol} < 1.0\text{ mol} (the initial amount introduced), liquid-vapour equilibrium is indeed established, leaving 1.0−0.2125=0.7875 mol1.0 - 0.2125 = 0.7875\text{ mol} of the compound in the liquid state (which occupies negligible volume).

Thus, the total volume of the gases is close to 7 L7\text{ L}.

PreviousNext