All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: To study a key fuel-cell reaction, a chemical engineer has 20.0-L tanks of H 2 and of O2 and wants to use up both tanks to form 28.0 mol of water at 23.8°C. (b) Use the van der Waals equation to find

Problem

To study a key fuel-cell reaction, a chemical engineer has 20.0-L tanks of H 2 and of O2 and wants to use up both tanks to form 28.0 mol of water at 23.8°C. 

(b) Use the van der Waals equation to find the pressure needed in each tank. 


Solution

The Van der Waals' Equation is as follows:

where a is the polarity coefficient, which accounts for the attractive or repulsive interactions of the gas molecules, and b is the size coefficient, which accounts for the size of the individual gas molecules. This equation show how real gases deviate from ideal behavior.

Let P + n2a/V2Pvdw, the van der Waals pressure. We can rearrange this equation to get Pvdw:

The reaction of H2 and O2 to form H2O is as follows:

2 H2(g) + O2(g) → 2 H2O(g)

We're given T = 23.8˚C + 273.15 = 296.95 K.

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