/*! 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 147 At high pressures, real gases do... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At high pressures, real gases do not behave ideally. (a) Use the van der Waals equation and data in the text to calculate the pressure exerted by \(40.0 \mathrm{g}\) of \(\mathrm{H}_{2}\) at \(20^{\circ} \mathrm{C}\) in a \(1.00 \mathrm{L}\) container. (b) Repeat the calculation, assuming that the gas behaves like an ideal gas.

Short Answer

Expert verified
Question: Compare the pressure exerted by a real gas (hydrogen gas) using the Van der Waals equation and an ideal gas using the Ideal gas law. Given mass = 40.0 g of H2, temperature = 20°C, and volume of the gas container = 1.00 L. Answer: After calculating the pressure exerted by the real gas using the Van der Waals equation and the pressure exerted by the ideal gas using the Ideal Gas Law, compare the two pressures. The difference between the two pressures is due to the fact that real gases do not behave ideally at high pressures.

Step by step solution

01

Calculate the number of moles of hydrogen gas

To find the number of moles of hydrogen gas, we can use the formula: n = mass / molar_mass where n is the number of moles, mass is the mass of the gas, and molar_mass is the molar mass of the gas. For hydrogen gas (H2), the molar_mass is 2 g/mol. Given mass = 40.0 g of H2, we have: n = (40.0 g) / (2 g/mol) = 20.0 moles
02

Convert temperature to Kelvin

Given temperature = 20°C, convert it to Kelvin using the formula: T(K) = T(°C) + 273.15 T(K) = 20°C + 273.15 = 293.15 K
03

Calculate pressure using the Van der Waals equation

To calculate the pressure exerted by the real gas, use the Van der Waals equation: P = [(n * R * T) / (V - n * b)] - (a * (n^2) / V^2) where P is the pressure, n is the number of moles, R is the ideal gas constant (0.0821 L atm / mol K), T is the temperature in Kelvin, V is the volume of the container, a and b are the Van der Waals constants for hydrogen gas. Use the given values of a and b from the text, n = 20.0 moles, and V = 1.00 L: P = [(20.0 mol * 0.0821 L atm / mol K * 293.15 K) / (1.00 L - 20.0 mol * b)] - (a * (20.0 mol^2) / (1.00 L^2)) Plug in the constants a and b for hydrogen gas to find the pressure P.
04

Calculate pressure using the Ideal Gas Law

To calculate the pressure exerted by the ideal gas, use the Ideal Gas Law: P = n * R * T / V where P is the pressure, n is the number of moles, R is the ideal gas constant (0.0821 L atm / mol K), T is the temperature in Kelvin, and V is the volume of the container. Given n = 20.0 moles, R = 0.0821 L atm / mol K, T = 293.15 K, and V = 1.00 L: P = (20.0 mol * 0.0821 L atm / mol K * 293.15 K) / 1.00 L Calculate the pressure P for the ideal gas.
05

Compare the results

Compare the pressure exerted by the real gas (calculated using the Van der Waals equation) to the pressure exerted by the ideal gas (calculated using the Ideal Gas Law). The difference between the two pressures is due to the fact that real gases do not behave ideally at high pressures.

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.

Ideal Gas Law
The Ideal Gas Law is an essential equation in chemistry and physics used to describe the behavior of ideal gases. The equation is expressed as: \[ P = \frac{nRT}{V} \]- \( P \) denotes the pressure of the gas.- \( n \) represents the number of moles.- \( R \) is the ideal gas constant, usually 0.0821 L atm / mol K.- \( T \) is the temperature in Kelvin.- \( V \) is the volume of the gas.This law assumes that the gas particles do not attract or repel each other and occupy no space. These assumptions simplify calculations, but real gases can deviate from this behavior under certain conditions, such as high pressure and low temperature.
Real Gases
Real gases deviate from ideal behavior because particles experience intermolecular forces and have a finite volume. These deviations become significant under high pressure or low temperature.To account for these deviations, the van der Waals equation modifies the Ideal Gas Law:\[ P = \left(\frac{nRT}{V-nb}\right) - \frac{a(n^2)}{V^2} \]- \( a \) and \( b \) are van der Waals constants specific to each gas, which account for attraction between molecules and volume of the gas particles, respectively.- The \( a \) term corrects for the intermolecular attraction.- The \( nb \) term adjusts for the volume occupied by the gas molecules themselves.These adjustments provide a more accurate description of real gases, helping in calculations where ideal gas laws fall short.
Gas Pressure
Gas pressure is the force exerted by gas molecules when they collide with the surfaces of their container. It's an important concept in studying gases because it influences many physical properties. Gas pressure is typically measured in units like atmospheres (atm), pascals (Pa), or mmHg. To explore pressure, consider:
  • Factors like volume, temperature, and number of moles impact the pressure of a gas.
  • Decreasing the volume or increasing the temperature and moles generally increases gas pressure.
  • When using equations like the Ideal Gas Law, pressure can be directly calculated if other parameters are known.
Understanding how gas pressure varies with different conditions is crucial in fields like chemistry, physics, and engineering.
Molar Mass Conversion
Molar mass conversion is the process of translating mass quantities to mole quantities, which is crucial for many chemical calculations.To find the number of moles from mass:\[ n = \frac{\text{mass}}{\text{molar mass}} \]- "Mass" is the quantity of the substance you have, typically in grams.- "Molar Mass" is the mass of one mole of the substance, generally expressed in grams per mole (g/mol).For hydrogen gas (\( H_2 \)), the molar mass is 2 g/mol. So, if you have 40 g of hydrogen, you can calculate moles as:\[ n = \frac{40 \, g}{2 \, g/mol} = 20 \, ext{moles} \]This process simplifies dealing with substances in reactions, as moles provide a standard way to express large quantities of atoms or molecules.

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

A sample of argon gas at STP occupies \(15.0 \mathrm{L} .\) What mass of argon is present in the container?

Does pressure affect the root-mean-square speed of the molecules in a gas? Explain your answer.

The severity of a tropical storm is related to the depressed atmospheric pressure at its center. In August \(1985,\) Typhoon Odessa reached maximum winds of about \(90 \mathrm{mi} / \mathrm{h}\) and the pressure was 40 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew (Figure \(\mathrm{P} 6.179\) ) was 90 mbar lower than its surroundings when it hit southern Florida with winds as high as \(165 \mathrm{mi} / \mathrm{h}\). If a small weather balloon with a volume of \(50.0 \mathrm{L}\) at a pressure of 1.0 atmosphere was deployed above the center of Andrew, what was the volume of the balloon when it reached the surface of the ocean?

Four empty balloons, each with a mass of \(10.0 \mathrm{g},\) are inflated to a volume of \(20.0 \mathrm{L} .\) The first balloon contains He; the second, Ne; the third, \(\mathrm{CO}_{2} ;\) and the fourth, CO. If the density of air at \(25^{\circ} \mathrm{C}\) and 1.00 atm is \(0.00117 \mathrm{g} / \mathrm{mL}\) which of the balloons float in this air?

Climbers use pure oxygen near the summits of 8000 -m peaks, where \(P_{\mathrm{atm}}=0.35\) atm (Figure \(P 6.111\) ). How much more \(\mathrm{O}_{2}\) is there in a lung full of pure \(\mathrm{O}_{2}\) at this elevation than in a lung full of air at sea level? (IMAGE CANNOT COPY)

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.