/*! 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 68 The size of an oxygen molecule i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The size of an oxygen molecule is about 2.0 \(\times 10^{-10} \mathrm{m}\) Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures \((T=300 \mathrm{K})\) .

Short Answer

Expert verified
The pressure at which the volume of the molecules causes noticeable deviations is when \( P \) is high enough that \( V \approxeq nb \).

Step by step solution

01

Understanding the van der Waals correction

For real gases, deviations from ideal-gas behavior are accounted for using van der Waals equation. The correction for volume accounts for the finite size of gas molecules, which is given by the term \( b \) in the equation. This volume correction is essential to estimate pressure deviations.
02

Determining the volume correction parameter \( b \)

The volume correction \( b \) roughly corresponds to four times the volume of a single molecule, calculated from its effective size. For an oxygen molecule with diameter \( d = 2.0 \times 10^{-10} \text{ m} \), calculate the volume of one molecule \( V_m \approx \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 \). Double this value is used for the van der Waals correction \( b = 4N_a V_m \), where \( N_a \) is Avogadro's number.
03

Calculating molecular volume

Calculate the volume of one oxygen molecule using the formula for the volume of a sphere: \[ V_m = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 \] Substituting \( d = 2.0 \times 10^{-10} \text{ m} \), obtain \( V_m \).
04

Estimating \( b \) for one mole of gas

Use the molecular volume calculated in the previous step to estimate \( b \) as \( b = 4 N_a V_m \). Use the value of Avogadro's number \( N_a = 6.022 \times 10^{23} \text{ molecules/mol} \) to find \( b \).
05

Applying van der Waals Pressure Correction

The deviations in pressure are meaningful when the term \( n^2a/V^2 \) becomes significant in the van der Waals equation \[ \left( P + \frac{n^2a}{V^2} \right) (V - nb) = nRT \] where \( a \) is a constant related to intermolecular forces. At conditions where molecular volume \( b \) is significant, \( P \) can be estimated roughly balancing these terms.
06

Estimating the threshold pressure

Assume when the volume occupied by the gas equals \( nb \), deviation is significant. Use \( V = nb \) in \( P = \frac{nRT}{V-nb} \). Substitute \( b \) obtained earlier and solve for \( P \) noting that \( T = 300 \text{ K} \).

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 a fundamental principle in chemistry and physics used to describe the behavior of gases. It is expressed as \( PV = nRT \), where:
  • \( P \) stands for pressure.
  • \( V \) is the volume of the gas.
  • \( n \) represents the number of moles.
  • \( R \) is the ideal gas constant.
  • \( T \) is the temperature in Kelvin.
This equation assumes that gases consist of non-interacting point particles with no volume and no intermolecular forces.
It is widely used because it simplifies calculations for an ideal scenario. However, real gases deviate from this behavior under high pressure or low temperature.
Understanding these deviations leads us to more complex models, such as the van der Waals equation, which account for real-world behaviors.
Molecular Volume
Molecular volume is the space occupied by a single molecule of gas.
In the context of the van der Waals equation, this concept is crucial to account for the finite size of gas molecules, which the Ideal Gas Law overlooks. To calculate molecular volume, consider the molecule's effective diameter. For a spherical molecule, its volume \( V_m \) is determined using the formula for a sphere:
  • \( V_m = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 \)
Here, \( d \) is the molecule's effective diameter.
For example, for an oxygen molecule with a diameter of \( 2.0 \times 10^{-10} \text{ m} \), you calculate \( V_m \) substituting \( d \) into the formula.
This helps in estimating the correction term \( b \) in van der Waals equation corresponding to four times the molecular volume for one mole of gas.
Real Gases
Real gases are gases that deviate from the behavior predicted by the Ideal Gas Law.
These deviations become especially notable at high pressures or low temperatures.
Unlike ideal gases, real gases have measurable volumes and experience intermolecular forces. To understand these deviations, the van der Waals equation is employed:
  • \( \left( P + \frac{n^2a}{V^2} \right) (V - nb) = nRT \)
Two main corrections differentiate it from the Ideal Gas Law:
  • The pressure term \( \frac{n^2a}{V^2} \), accounting for intermolecular attractions.
  • The volume term \( nb \), considering the volume occupied by molecules themselves.
The constants \( a \) and \( b \) are specific to each gas.
Real gases approximate ideal behavior only when intermolecular forces are negligible, usually at high temperatures and low pressures.
Pressure Deviation
Pressure deviation refers to the difference in pressure between real gas behavior and the predictions by the Ideal Gas Law.
Van der Waals equation helps in quantifying these deviations.In practical scenarios, real gases show deviations because of:
  • Intermolecular forces (adjusted by the parameter \( a \) in the van der Waals equation).
  • Finite molecular volume (adjusted by the parameter \( b \)).
To estimate when these deviations become significant, one can focus on the pressure term \( P + \frac{n^2a}{V^2} \).
This becomes relevant particularly when gas molecules take up a significant portion of the total volume or at critical conditions where attractions or repulsions are pronounced.
For oxygen, as calculated in the original exercise, deviations become noticeable around pressures where \( nb \) effectively impacts the volume consideration, leading to deviations from ideal behavior.

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 large tank of water has a hose connected to it, as shown in Fig. P18.65. The tank is sealed at the top and has compressed air between the water surface and the top. When the water height \(h\) has the value \(3.50 \mathrm{m},\) the absolute pressure \(p\) of the compressed air is \(4.20 \times\) \(10^{5}\) Pa. Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be \(1.00 \times 10^{5}\) Pa. (a) What is the speed with which water flows out of the hose when \(h=3.50 \mathrm{m} ?\) (b) As water flows out of the tank, \(h\) decreases. Calculate the speed of flow for \(h=3.00 \mathrm{m}\) and for \(h=2.00 \mathrm{m} .\) (c) At what value of \(h\) does the flow stop?

The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure, since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of 0.600 L at a temperature of \(19.0^{\circ} \mathrm{C} .\) What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen \((77.3 \mathrm{K}) ?\)

During a test dive in \(1939,\) prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature at the surface was \(27.0^{\circ} \mathrm{C},\) and at the bottom it was \(7.0^{\circ} \mathrm{C}\) . The density of seawater is 1030 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 \(\mathrm{m}\) high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

An automobile tire has a volume of 0.0150 \(\mathrm{m}^{3}\) on a cold day when the temperature of the air in the tire is \(5.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.02 atm. Under these conditions the gauge pressure is measured to be 1.70 atm (about 25 \(\mathrm{lb} / \mathrm{in.} .\) ). After the car is driven on the highway for 30 min, the temperature of the air in the tires has risen to \(45.0^{\circ} \mathrm{C}\) and the volume has risen to 0.0159 \(\mathrm{m}^{3}\) . What then is the gauge pressure?

Hydrogen on the Sun. The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times 10^{-27}\) kg. (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2},\) where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant \((\) see Example 13.5 of Section 13.3\() .\) Use the data in Appendix \(F\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

See all solutions

Recommended explanations on Physics 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.