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From the density of liquid water and its molar mass, calculate the volume that 1 mol liquid water occupies. If water were an ideal gas at STP, what volume would a mole of water vapor occupy? Can we achieve the STP conditions for water vapor? Why or why not?

Short Answer

Expert verified
1 mol of liquid water occupies 18.02 mL. Water vapor at STP would occupy ~22.4 L, but STP conditions are not feasible for water vapor as it condenses.

Step by step solution

01

Understanding the Problem

We are asked to calculate the volume occupied by 1 mole of liquid water and 1 mole of water vapor at STP. We also need to determine if the STP conditions can be achieved for water vapor.
02

Volume of Liquid Water

To find the volume of liquid water, we use the density formula: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). The density of liquid water is approximately 1.00 g/cm³, and the molar mass of water is 18.02 g/mol. Thus: \[ \text{Volume} = \frac{18.02\, \text{g/mol}}{1.00\, \text{g/cm}^3} = 18.02\, \text{cm}^3 \] or 18.02 mL.
03

Volume of Water Vapor under Ideal Gas Behavior

By assuming water vapor as an ideal gas at STP (Standard Temperature and Pressure: 0°C, 1 atm), we can use the ideal gas law equation: \( PV = nRT \) where \( n = 1 \) mol, \( R = 0.0821 \frac{L\cdot atm}{mol\cdot K} \), and \( T = 273.15 \ K \) (STP). Solving for \( V \): \[ V = \frac{nRT}{P} = \frac{(1\, mol)(0.0821\, L\cdot atm/mol\cdot K)(273.15\, K)}{1\, atm} \approx 22.4\, L \]
04

Feasibility of Achieving STP for Water Vapor

At STP, the temperature is 0°C and pressure is 1 atm. Water boils and turns to vapor at 100°C at 1 atm pressure, much higher than 0°C. Therefore, it is not feasible to have water vapor at STP as it would condense into liquid water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as \( PV = nRT \). This equation assumes the gas behaves ideally, meaning the particles do not interact with each other beyond elastic collisions, and occupy no volume themselves. Here,
  • \( P \) is the pressure of the gas in atm
  • \( V \) is the volume in liters
  • \( n \) is the number of moles of the gas
  • \( R \) is the ideal gas constant, approximately 0.0821 Lâ‹…atm/molâ‹…K
  • \( T \) is the temperature in Kelvin
To find the volume of one mole of any gas under standard temperature and pressure (STP, 0°C and 1 atm), we use the values in the ideal gas law equation:\[ V = \frac{nRT}{P} = \frac{(1\, mol)(0.0821\, L\cdot atm/mol\cdot K)(273.15\, K)}{1\, atm} = 22.4\, L \]This tells us that one mole of any ideal gas occupies roughly 22.4 liters at STP. However, real gases may deviate from this volume due to intermolecular forces and finite volume of particles.
Density of Water
Density is a measure of how much mass a substance contains within a specific volume, expressed as mass per unit of volume, often in g/cm³ for liquids. The density of water at room temperature is approximately 1.00 g/cm³. This means that every cubic centimeter of water contains one gram of mass.
To calculate the volume of one mole of water, we use the formula:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]Given that the molar mass of water is 18.02 g/mol, the volume of one mole of liquid water is:\[ \text{Volume} = \frac{18.02\, \text{g/mol}}{1.00\, \text{g/cm}^3} = 18.02\, \text{cm}^3 \]This equals 18.02 milliliters. Liquid water has a much smaller volume compared to water vapor as an ideal gas, which shows how dense the liquid state is compared to the gaseous state.
Standard Temperature and Pressure (STP)
Standard temperature and pressure (STP) are the conditions set for experiments to ensure consistency and comparability of results. STP is defined by a temperature of 0°C (273.15 K) and a pressure of 1 atm. At these conditions, the behavior of gases is often simplified and assumed to follow the ideal gas law closely. For water, creating vapor at STP is challenging. Water boils and transitions from liquid to vapor at 100°C under 1 atm, which is much higher than the 0°C specified by STP. If we lower the temperature to 0°C while maintaining 1 atm pressure, water vapor would likely condense back into liquid form. Therefore, achieving STP for water vapor is not feasible because water naturally exists in the liquid state at this pressure and temperature, showcasing the limits of idealized conditions when applied to real-world substances.

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Most popular questions from this chapter

Acetylene can be made by reacting calcium carbide with water. $$\mathrm{CaC}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})$$ Assume that you place \(2.65 \mathrm{~g} \mathrm{CaC}_{2}\) in excess water and collect the acetylene over water. The volume of the acetylene and water vapor is \(795 \mathrm{~mL}\) at \(25.0^{\circ} \mathrm{C}\) and a barometric pressure of \(735.2 \mathrm{mmHg}\). Calculate the percent yield of acetylene. The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8 \mathrm{mmHg}\).

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