Problem 43 - Chemistry and Chemical Reactivity

Question

If you place \(1.0 \mathrm{L}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in a room that is \(3.0 \mathrm{m}\) long, \(2.5 \mathrm{m}\) wide, and \(2.5 \mathrm{m}\) high, will all of the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethanol at \(25^{\circ} \mathrm{C}\) is \(59 \mathrm{mm} \mathrm{Hg},\) and the density of the liquid at this temperature is \(0.785 \mathrm{g} / \mathrm{cm}^{3}\)

Step-by-Step Solution

Answer

All the ethanol will evaporate as the available moles of ethanol are less than needed to saturate the air.

Step 1: Calculate Volume of Room

First, find the volume of the room in cubic meters. The room is a rectangular prism, so multiply its length, width, and height: \( V = 3.0 \, \text{m} \times 2.5 \, \text{m} \times 2.5 \, \text{m} = 18.75 \, \text{m}^3 \). Convert this volume to liters, knowing that 1 \( \text{m}^3 = 1000 \, \text{L} \): \( 18.75 \, \text{m}^3 = 18750 \, \text{L} \).

Step 2: Convert Vapor Pressure to Moles

Convert the vapor pressure of ethanol to atmospheres to use in further calculations: \( \frac{59 \, \text{mmHg}}{760 \, \text{mmHg} / \text{atm}} = 0.0776 \, \text{atm} \). Use the ideal gas law \( PV = nRT \) to find the number of moles of ethanol at this pressure. Assume the entire room is available for gas, \( V = 18750 \, \text{L} \), \( R = 0.0821 \, \text{L} \, \text{atm} / \text{mol} \, \text{K} \), and \( T = 298 \, \text{K} \) (as a reasonable room temperature estimate): \[ n = \frac{PV}{RT} = \frac{0.0776 \, \text{atm} \times 18750 \, \text{L}}{0.0821 \, \text{L} \, \text{atm} / \text{mol} \, \text{K} \times 298 \, \text{K}} = 59.48 \, \text{mol} \].

Step 3: Convert Liquid Ethanol to Moles

Convert the volume of liquid ethanol to grams using its density: \( 1.0 \, \text{L} = 1000 \, \text{cm}^3 \), so \( 1000 \, \text{cm}^3 \times 0.785 \, \text{g/cm}^3 = 785 \, \text{g} \). Next, convert grams to moles using the molar mass of ethanol, \( \text{C}_2\text{H}_5\text{OH} \approx 46.07 \, \text{g/mol} \): \( \frac{785 \, \text{g}}{46.07 \, \text{g/mol}} = 17.04 \, \text{mol} \).

Step 4: Compare Moles of Ethanol

Compare the calculated moles of ethanol that can evaporate \( (59.48 \, \text{mol}) \) to the amount you initially have \( (17.04 \, \text{mol}) \). Since the moles of ethanol initially available are less than the calculated amount needed to saturate the air, all the ethanol will evaporate.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation used to relate the pressure, volume, temperature, and number of moles of a gas. It is often expressed as \( PV = nRT \), where:

• \(P\) is the pressure of the gas in atmospheres (atm),
• \(V\) is the volume of the gas in liters (L),
• \(n\) is the number of moles of the gas,
• \(R\) is the ideal gas constant (approximately \(0.0821 \, \text{L} \, \text{atm} / \text{mol} \, \text{K}\)), and
• \(T\) is the temperature in Kelvin (K).

To use this equation, ensure that all your units are correct and consistent. You need to convert pressure from mmHg to atm and volume into liters if they are not already, as these are the units compatible with the gas constant \(R\). The equation enables you to determine one of the variables if the others are known.

Moles Calculation

When dealing with chemical quantities, the concept of moles is crucial. It allows chemists to quantify substances at the molecular level. For gases, the Ideal Gas Law is applicable to calculate moles when pressure, volume, and temperature are known. Here’s a step-by-step method used in the exercise:

• First, determine the volume of space available for the gas, converting all measurements to liters.
• Next, adjust the given pressure to atm as needed.
• Temperature should be converted to Kelvin by adding 273.15 to the Celsius value.

Using the calculations in the exercise, you can find the number of moles (\(n\)) by rearranging the Ideal Gas Law to \(n = \frac{PV}{RT}\). This calculation tells us the maximum number of moles of ethanol that can vaporize within the space provided.

Ethanol Evaporation

Ethanol evaporation is essentially the process of ethanol transitioning from a liquid state to a gaseous state. This occurs until the air reaches saturation at a given temperature and pressure. The amount of ethanol that can evaporate depends on the vapor pressure and available volume. In the presented exercise:

• We calculate the theoretical maximum moles of ethanol using its given vapor pressure and the room volume.
• Compare these calculated moles to the starting moles, based on the amount of liquid ethanol initially present.

Since the calculated moles that can fit in the gas phase (59.48 mol) are greater than the moles present in liquid form (17.04 mol), we deduce that all available ethanol evaporates. Thus, at room conditions, there won’t be any remaining liquid if there's enough room volume.

Density and Molar Mass Calculations

Density and molar mass are two critical concepts that help relate different properties of a substance. Density is the mass per unit volume, which is used to convert between volume and mass. Molar mass, meanwhile, is the mass of one mole of a substance, and is essential for converting between mass and moles.Here’s how these calculations work in unity:

• First, use the density to convert the ethanol volume (\(1.0 \, \text{L}\) in this case) into mass using \(\text{density} = \frac{\text{mass}}{\text{volume}}\).
• Multiply the volume by density to find the mass (785 g from the problem).
• Then divide this mass by the molar mass of ethanol \(\approx 46.07 \, \text{g/mol}\) to determine the moles of ethanol available.

This interconnected use of density and molar mass allows precise calculations of how much substance you have and what can occur in terms of phase transitions in a saturated environment.