(a)
Interpretation:
Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.
Concept introduction:
The
Answer to Problem 1.79E
The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;
Van der Waals constant for Krypton a = 2.318 atm L2/mol2;
b = 0.03978 L/mol
Boyle temperature,
Molar volume for krypton
Explanation of Solution
The non-ideal gas equation represented as;
In the above equation,
[p+ an2/V2] Correction term introduced for molecular attraction
[V– nb] correction term introduced for volume of molecules
‘a’ and ‘b’ are called as van der Waals constants
a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.
b = the volume correction and it is having relationship to the size of the gas particles.
Given;
Van der Waals constant for Krypton a =
Boyle temperature
=
= 711.04 K
At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure
Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.
(b)
Interpretation:
Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C2H6 is to be calculated at 25 °C and 1 bar pressure.
Concept introduction:
The ideal
Answer to Problem 1.79E
The molar volumes of ethane, C2H6 is calculated at 25 °C and 1 bar pressure as follows;
Van der Waals constant for ethane a = 5.489 atm L2/mol2;
b = 0.0638 L/mol
Boyle temperature Tb = a/bR = 1049.5 K
Molar volume for ethane ῡ = RT/p = 87.2 L
Explanation of Solution
The non-ideal gas equation represented as;
In the above equation,
[p + an2/V2] Correction term introduced for molecular attraction
[V – nb] Correction term introduced for volume of molecules
‘a’ and ‘b’ are called as van der Waals constants
a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.
b = the volume correction and it is having relationship to the size of the gas particles.
Given;
Van der Waals constant for ethane a = 5.489 atm L2/mol2
b = 0.0638 L/mol
Boyle temperature Tb = a/bR
At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure
ῡ = RT/p
Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C2H6 is calculated at 25 °C and 1 bar pressure.
(c)
Interpretation:
Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.
Concept introduction:
The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.
Answer to Problem 1.79E
The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;
Van der Waals constant for mercury a = 8.093atm L2/mol2;
b = 0.01696 L/mol
Boyle temperature Tb = a/bR = 5822 K
Molar volume for mercury ῡ = RT/p = 484 L
Explanation of Solution
The non-ideal gas equation represented as;
In the above equation,
[p + an2/V2] Correction term introduced for molecular attraction
[V – nb] Correction term introduced for volume of molecules
‘a’ and ‘b’ are called as van der Waals constants
a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.
b = the volume correction and it is having relationship to the size of the gas particles.
Given;
Van der Waals constant for mercury a = 8.093atm L2/mol2;
b = 0.01696 L/mol
Boyle temperature Tb = a/bR
= (8.093 atm L2 mol-2)/(0.01696 L mol-1 x 0.08205 L. atm K-1 mol-1
= 5822 K
At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure
ῡ = RT/p
Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.
Want to see more full solutions like this?
Chapter 1 Solutions
PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
- Do the Lone Pairs get added bc its valence e's are a total of 6 for oxygen and that completes it or due to other reasons. How do we know the particular indication of such.arrow_forwardNGLISH b) Identify the bonds present in the molecule drawn (s) above. (break) State the function of the following equipments found in laboratory. Omka) a) Gas mask b) Fire extinguisher c) Safety glasses 4. 60cm³ of oxygen gas diffused through a porous hole in 50 seconds. How long w 80cm³ of sulphur(IV) oxide to diffuse through the same hole under the same conditions (S-32.0.0-16.0) (3 m 5. In an experiment, a piece of magnesium ribbon was cleaned with steel w clean magnesium ribbon was placed in a crucible and completely burnt in oxy cooling the product weighed 4.0g a) Explain why it is necessary to clean magnesium ribbon. Masterclass Holiday assignmen PB 2arrow_forwardHi!! Please provide a solution that is handwritten. Ensure all figures, reaction mechanisms (with arrows and lone pairs please!!), and structures are clearly drawn to illustrate the synthesis of the product as per the standards of a third year organic chemistry course. ****the solution must include all steps, mechanisms, and intermediate structures as required. Please hand-draw the mechanisms and structures to support your explanation. Don’t give me AI-generated diagrams or text-based explanations, no wordy explanations on how to draw the structures I need help with the exact mechanism hand drawn by you!!! I am reposting this—ensure all parts of the question are straightforward and clear or please let another expert handle it thanks!!arrow_forward
- In three dimensions, explain the concept of the velocity distribution function of particles within the kinetic theory of gases.arrow_forwardIn the kinetic theory of gases, explain the concept of the velocity distribution function of particles in space.arrow_forwardIn the kinetic theory of gases, explain the concept of the velocity distribution function of particles.arrow_forward
- Hi!! Please provide a solution that is handwritten. this is an inorganic chemistry question please answer accordindly!! its just one question with parts JUST ONE QUESTION with its parts spread out till part (g), please answer EACH part till the end and dont just provide wordy explanations wherever asked for structures, please DRAW DRAW them on a paper and post clearly!! answer the full question with all calculations step by step EACH PART CLEARLY please thanks!! im reposting this please solve all parts and drawit not just word explanations!!arrow_forwardHi!! Please provide a solution that is handwritten. this is an inorganic chemistry question please answer accordindly!! its just one question with parts JUST ONE QUESTION, please answer EACH part PART A AND PART B!!!!! till the end and dont just provide wordy explanations wherever asked for structures, please DRAW DRAW them on a paper and post clearly!! answer the full question with all details EACH PART CLEARLY please thanks!! im reposting this please solve all parts and drawit not just word explanations!!arrow_forwardHi!! Please provide a solution that is handwritten. this is an inorganic chemistry question please answer accordindly!! its just one question with parts JUST ONE QUESTION, please answer EACH part till the end and dont just provide wordy explanations wherever asked for structures, please DRAW DRAW them on a paper and post clearly!! answer the full question with all details EACH PART CLEARLY please thanks!! im reposting this please solve all parts and drawit not just word explanations!!arrow_forward
- 8b. Explain, using key intermediates, why the above two products are formed instead of the 1,2-and 1,4- products shown in the reaction below. CIarrow_forward(5pts) Provide the complete arrow pushing mechanism for the chemical transformation depicted below Use proper curved arrow notation that explicitly illustrates all bonds being broken, and all bonds formed in the transformation. Also, be sure to include all lone pairs and formal charges on all atoms involved in the flow of electrons. CH3O H I I CH3O-H H I ① Harrow_forward6. Draw the products) formed from the following reactions. (a) HIarrow_forward
- Chemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningPhysical ChemistryChemistryISBN:9781133958437Author:Ball, David W. (david Warren), BAER, TomasPublisher:Wadsworth Cengage Learning,