/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A steel cylinder holds \(1.50 \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A steel cylinder holds \(1.50 \mathrm{g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). What is the pressure of the ethanol vapor if the cylinder has a volume of \(251 \mathrm{cm}^{3}\) and the temperature is \(250^{\circ} \mathrm{C} ?\) (Assume all of the ethanol is in the vapor phase at this temperature.)

Short Answer

Expert verified
The pressure of the ethanol vapor is approximately 5.56 atm.

Step by step solution

01

Convert mass to moles

Find the molar mass of ethanol, \(\text{C}_2\text{H}_5\text{OH}\). The molar mass is calculated as \((2 \times 12.01) + (6 \times 1.01) + 16.00 = 46.08\, \text{g/mol}\). Given \(1.50\, \text{g}\) of ethanol, calculate the number of moles, using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Thus, \(\text{moles} = \frac{1.50}{46.08} \approx 0.0325\, \text{mol}\).
02

Convert temperature to Kelvins

Convert the temperature from Celsius to Kelvin using the formula: \(T(K) = T(\degree C) + 273.15\). So, \(250^{\circ} \text{C} = 250 + 273.15 = 523.15\, \text{K}\).
03

Convert volume to liters

Convert the volume from cubic centimeters to liters. Since \(1\, \text{L} = 1000\, \text{cm}^3\), we have \(V = 251\, \text{cm}^3 \times \frac{1\, \text{L}}{1000\, \text{cm}^3} = 0.251\, \text{L}\).
04

Use the ideal gas law

Use the ideal gas law to find the pressure. The ideal gas law is given by \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the ideal gas constant \(0.0821\, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}\), and \(T\) is temperature in Kelvin. Rearranging for \(P\), we get \(P = \frac{nRT}{V}\).
05

Calculate pressure

Substitute the known values into the equation: \(P = \frac{0.0325 \times 0.0821 \times 523.15}{0.251}\). This simplifies to \(P \approx \frac{1.396}{0.251} \approx 5.56\, \text{atm}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Ethanol Vapor Pressure
Understanding ethanol vapor pressure is essential when studying its behavior in different conditions. Ethanol is a colorless liquid that can evaporate very quickly, creating vapor—an essential property in chemistry.

Ethanol vapor pressure refers to the pressure exerted by the ethanol molecules in the vapor phase above the liquid. It's important because it helps determine how easily ethanol will evaporate at a given temperature. As temperature increases, so does the kinetic energy of ethanol molecules, raising the vapor pressure.
  • At higher temperatures, ethanol has higher vapor pressures, meaning it evaporates more quickly.
  • Vapor pressure is a key factor in applications like distillation and the food industry.
In the original exercise, understanding ethanol vapor pressure helps us use the ideal gas law to calculate the pressure it creates inside a container, aiding in essential calculations for chemical processes.
Molar Mass Calculation
Calculating the molar mass of a compound is a foundational skill in chemistry. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

For ethanol (C_2H_5OH), the molar mass is calculated by summing the atomic masses of all atoms in the molecule:
  • Carbon (C): 2 atoms × 12.01 g/mol = 24.02 g/mol
  • Hydrogen (H): 6 atoms × 1.01 g/mol = 6.06 g/mol
  • Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol
Adding these together, ethanol's molar mass is 46.08 g/mol. This value allows us to convert the mass of ethanol into moles using the formula: \( ext{moles} = rac{ ext{mass}}{ ext{molar mass}}\). It's a crucial step in applying the ideal gas law, as it helps calculate the amount of substance present.
Temperature Conversion to Kelvin
Temperature conversion is an important aspect when dealing with gas laws and chemistry experiments. The Kelvin scale is the SI unit for temperature and provides an absolute reference point.

Converting Celsius to Kelvin is straightforward:
  • Add 273.15 to the Celsius temperature.
For the exercise given:
  • \(250^{ ext{°C}} + 273.15 = 523.15 ext{K}\),
Converting to Kelvin is necessary because the ideal gas law, \(PV = nRT\), requires the temperature to be in Kelvin to ensure all calculations are consistent. Kelvin provides a direct measure of kinetic energy measures—key to understanding how temperature affects gas behavior.
Volume Conversion to Liters
Volume conversion is another key element in chemical calculations. Many substances are measured in cubic centimeters (cm³) but chemists often need this in liters (L) for calculations using the ideal gas law.

To convert from cm³ to L:
  • Remember that 1 L = 1000 cm³.
  • Simply divide the volume in cm³ by 1000.
In the given exercise:
  • 251 cm³ becomes \( 0.251 ext{ L}\) after dividing by 1000.
This conversion aligns unit measurements needed for gas law calculations, ensuring consistency and accuracy when using the equation \(PV = nRT\). Proper conversion prevents errors and facilitates smooth computations in a scientific context.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If the absolute temperature of a gas doubles, by how much does the rms speed of the gaseous molecules increase?

Acetaldehyde is a common liquid compound that vaporizes readily. Determine the molar mass of acetaldehyde from the following data: Sample mass \(=0.107 \mathrm{g} \quad\) Volume of gas \(=125 \mathrm{mL}\) Temperature \(=0.0^{\circ} \mathrm{C} \quad\) Pressure \(=331 \mathrm{mm} \mathrm{Hg}\)

A You have a \(550 .\) -mL. tank of gas with a pressure of 1.56 atm at \(24^{\circ} \mathrm{C} .\) You thought the gas was pure carbon monoxide gas, CO, but you later found it was contaminated by small quantities of gaseous \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\). Analysis shows that the tank pressure is 1.34 atm (at \(24^{\circ} \mathrm{C}\) ) if the \(\mathrm{CO}_{2}\) is removed. Another experiment shows that \(0.0870 \mathrm{g}\) of \(\mathrm{O}_{2}\) can be removed chemically. What are the masses of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) in the tank, and what is the partial pressure of each of the three gases at \(25^{\circ} \mathrm{C} ?\)

The sodium azide required for automobile air bags is made by the reaction of sodium metal with dinitrogen monoxide in liquid ammonia: $$\begin{aligned} 3 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+4 \mathrm{Na}(\mathrm{s})+\mathrm{NH}_{3}(\ell) & \rightarrow \\ & \mathrm{NaN}_{3}(\mathrm{s})+3 \mathrm{NaOH}(\mathrm{s})+2 \mathrm{N}_{2}(\mathrm{g}) \end{aligned}$$ (a) You have \(65.0 \mathrm{g}\) of sodium, a \(35.0-\mathrm{L}\). flask containing \(\mathrm{N}_{2} \mathrm{O}\) gas with a pressure of 2.12 atm at \(23^{\circ} \mathrm{C}\) and excess ammonia. What is the theoretical yield (in grams) of NaNg? (b) Draw a Lewis structure for the azide ion. Include all possible resonance structures. Which resonance structure is most likely? (c) What is the shape of the azide ion?

\(\mathrm{Ni}(\mathrm{CO})_{4}\) can be made by reacting finely divided nickel with gaseous CO. If you have CO in a \(1.50-\mathrm{L}\). flask at a pressure of \(418 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C},\) along with \(0.450 \mathrm{g}\) of Ni powder, what is the theoretical yield of \(\mathrm{Ni}(\mathrm{CO})_{4} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.