EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106906149
Author: Jewett
Publisher: Cengage Learning US
bartleby

Videos

Textbook Question
Book Icon
Chapter 20, Problem 21P

Air (a diatomic ideal gas) at 27.0°C and atmospheric pressure is drawn into a bicycle pump that has a cylinder with an inner diameter of 2.50 cm and length 50.0 cm. The downstroke adiabatically compresses the air, which readies a gauge pressure of 8.00 × 105 Pa before entering the tire. We wish to investigate the temperature increase of the pump. (a) What is the initial volume of the air in the pump? (b) What is the number of moles of air in the pump? (c) What is the absolute pressure of the compressed air? (d) What is the volume of the compressed air? (c) What is the temperature of the compressed air? (f) What is the increase in internal energy of the gas during the compression? What If? The pump is made of steel that is 2.00 mm thick. Assume 4.00 cm of the cylinder’s length is allowed to come to thermal equilibrium with the air. (g) What is the volume of steel in this 4.00-cm length? (h) What is the mass of steel in this 4.00-cm length? (i) Assume the pump is compressed once. After the adiabatic expansion, conduction results in the energy increase in part (f) being shared between the gas and the 4.00-cm length of steel. What will be the increase in temperature of the steel after one compression?

(a)

Expert Solution
Check Mark
To determine

The initial volume of the air in the pump.

Answer to Problem 21P

The initial volume of the air in the pump is 2.45×104m3.

Explanation of Solution

Initial temperature for diatomic gasis 27.0°C, length of the pump is 50.0cm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

Write the expression to calculate the radius of the pump.

r=d2

Here,

r is the radius of the pump.

d is the diameter of the pump.

Write the formula to calculate the initial volume of the air in the pump.

V1=πr2l (1)

Here,

V1 is the initial volume of the air in the pump.

l is the length of the pump.

Substitute d2 for r in equation (1) to find V1,

    V1=π(d2)2l (2)

Substitute 2.50cm for d, 50.0cm for l in equation (2) to find V1,

    V1=π(2.50cm×(1m100cm)2)2(50.0cm×1m100cm)=2.45×104m3

Thus, the initial volume of the air in the pump is 2.45×104m3.

Conclusion:

Therefore, the initial volume of the air in the pump is 2.45×104m3.

(b)

Expert Solution
Check Mark
To determine

The number of moles of air in the pump.

Answer to Problem 21P

The number of moles of air in the pump is 9.97×103moles.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

Write the formula to calculate the number of moles of air in the pump.

    P1V1=nRT1n=P1V1RT1        (3)

Here,

n is the number of moles of air in the pump.

P1 is the atmospheric pressure for diatomic gas.

R is the ideal gas constant.

T1 is the initial temperature for diatomic gas.

The value of atmospheric pressure for diatomic gas is 1.013×105Pa and the value of ideal gas constant is 8.314J/molK.

Substitute 2.45×104m3 for V1, 1.013×105Pa for P1, 8.314J/molK for R, 27.0°C for T1 in equation (3) to find n,

    n=(1.013×105Pa)×(2.45×104m3)(8.314J/molK)×(27.0°C+273)K=9.97×103moles

Thus, the number of moles of air in the pump is 9.97×103moles.

Conclusion:

Therefore, the number of moles of air in the pump is 9.97×103moles.

(c)

Expert Solution
Check Mark
To determine

The absolute pressure of the compressed air.

Answer to Problem 21P

The absolute pressure of the compressed air is 9.01×105Pa.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

Write the formula to calculate the absolute pressure of the compressed air.

    P2=Pg+P1 (4)

Here,

P2 is the absolute pressure of the compressed air.

Pg is the gauge pressure for the compressed air.

Substitute 8.00×105Pa for Pg, 1.013×105Pa for P1 in equation (4) to find P2,

    P2=8.00×105Pa+1.013×105Pa=9.013×105Pa9.01×105Pa

Thus, the absolute pressure of the compressed air is 9.01×105Pa.

Conclusion:

Therefore, the absolute pressure of the compressed air is 9.01×105Pa.

(d)

Expert Solution
Check Mark
To determine

The volume of the compressed air.

Answer to Problem 21P

The volume of the compressed air is 5.15×105m3.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

Write the expression for the adiabatic compression.

    P1V1γ=P2V2γ

Here,

V2 is the volume of the compressed air.

γ is the specific heat ratio.

Write the formula to calculate the volume of the compressed air.

    V2=V1(P1P2)1γ (5)

Substitute 9.01×105Pa for P2, 1.013×105Pa for P1, 2.45×104m3 for V1, 1.40 for γ in equation (5) to find V2,

    V2=(2.45×104m3)(1.013×105Pa9.01×105Pa)11.40=5.145×105m35.15×105m3

Thus, the volume of the compressed air is 5.15×105m3.

Conclusion:

Therefore, the volume of the compressed air is 5.15×105m3.

(e)

Expert Solution
Check Mark
To determine

The temperature of the compressed air.

Answer to Problem 21P

The temperature of the compressed air is 560K.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

Write the formula to calculate the temperature of the compressed air.

    P2V2=nRT2T2=P2V2nR        (6)

Here,

T2 is the temperature of the compressed air.

Substitute 5.15×105m3 for V2, 9.013×105Pa for P2, 8.314J/molK for R, 9.97×103moles for n in equation (6) to find T2,

    T2=P2V2nRT2=(9.013×105Pa)×(5.15×105m3)(9.97×103moles)×(8.314J/molK)=559.978K560K

Thus, the temperature of the compressed air is 560K.

Conclusion:

Therefore, the temperature of the compressed air is 560K.

(f)

Expert Solution
Check Mark
To determine

The increase in internal energy of the gas during the compression.

Answer to Problem 21P

The increase in internal energy of the gas during the compression is 53.9J.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

For adiabatic process, the work done on the gas is equal to the change in internal energyof the gas during the compression.

    W=ΔU (7)

Here,

W is the work done on the gas.

ΔU is the change in internal energyof the gas during the compression.

Write the expression for the change in internal energyof the gas during the compression.

    ΔU=nCVΔT (8)

Here,

CV is the specific heat at constant volume.

Write the expression for specific heat at constant volume.

    CV=52R (9)

Here,

CV is the specific heat at constant volume.

Equate the three expressions (7),(8) and (9)and re-arrange to get ΔU,

    ΔU=n(52R)ΔT (10)

Write the formula to calculate the change in temperature of a monatomic ideal gas.

    ΔT=T2T1 (11)

Here,

ΔT is the change in temperature of a monatomic ideal gas.

Substitute 560K for T2, 300K for T1 in equation (11) to find ΔT,

    ΔT=T2T1=560K300K=260K

Thus, the change in temperature of a monatomic ideal gas is 260K.

Substitute 260K for ΔT, 8.314J/molK for R, 9.97×103moles for n in equation (10) to find ΔU,

    ΔU=(9.97×103moles)(52×8.314J/molK)×260K=53.87J53.9J

Thus, the increase in internal energy of the gas during the compression is 53.9J.

Conclusion:

Therefore, the increase in internal energy of the gas during the compression is 53.9J.

(g)

Expert Solution
Check Mark
To determine

The volume of the steel in this 4.00cm length.

Answer to Problem 21P

The volume of the steel in this 4.00cm length is 6.79×106m3.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, length of the cylinder is 4.00cm, thickness of the pump is 2.00mm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

Write the formula to calculate the volume of the steel in this 4.00cm length.

    Vs=πrs2l (12)

Here,

rs is the radius of the pump for steel.

Vs is the volume of the steel in this 4.00cm length.

Write the expression to calculate the square radius of the pump for steel.

    rs2=r22(d2)2 (13)

Here,

r2 is the outer radius of the pump.

Write the formula to calculate the outer radius of the pump.

    r2=d2+t (14)

Here,

t is the thickness of the pump.

Substitute 2.00mm for t, 2.50cm for r in equation (14) to find r2,

    r2=(2.50cm2×10mm1cm)+(2.00mm)=14.5mm

Thus, the outer radius of the pump is 14.5mm.

Substitute 14.5mm for r2, 2.50cm for r in equation (13) to find rs,

    rs2=r22(d2)2=(14.5mm)2(2.5cm×10mm1cm2)2=54mm

Thus, the square radius of the pump for steel is 54mm.

Substitute 54mm for rs2, 4.00cm for l in equation (12) to find Vs,

    Vs=π×(54mm×1m1000mm)×(4.00cm×1m100cm)=6.785×106m36.79×106m3

Thus, the volume of the steel in this 4.00cm length is 6.79×106m3.

Conclusion:

Therefore, the volume of the steel in this 4.00cm length is 6.79×106m3.

(h)

Expert Solution
Check Mark
To determine

The mass of the steel in this 4.00cm length.

Answer to Problem 21P

The mass of the steel in this 4.00cm length is 53.3g.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, length of the cylinder is 4.00cm, thickness of the pump is 2.00mm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

Write the formula to calculate the mass of the steel in this 4.00cm length.

    ρs=mVsm=ρs×Vs (15)

Here,

m is the mass of the steel in this 4.00cm length.

ρs is the density of the steel.

The value of density of the steel is 7.85×106g/m3.

Substitute 7.85×106g/m3 for ρs, 6.79×106m3 for Vs, in equation (15) to find m,

    m=(7.85×106g/m3)×(6.79×106m3)=53.3g

Thus, the mass of the steel in this 4.00cm length is 53.3g.

Conclusion:

Therefore, the mass of the steel in this 4.00cm length is 53.3g.

(i)

Expert Solution
Check Mark
To determine

The increase in temperature of the steel after one compression.

Answer to Problem 21P

The increase in temperature of the steel after one compression is 2.24K.

Explanation of Solution

Initial temperature for diatomic gas is 27.0°C, length of the pump is 50.0cm, length of the cylinder is 4.00cm, thickness of the pump is 2.00mm, diameter of the pump is 2.50cm, gauge pressure for the compressed air is 8.00×105Pa.

After the adiabatic compression, conduction in the part (f) being shared between the gas and the 4.00cm length of steel.

The work done on the gas is equal to the sum of change in internal energyof the gas during the compression and the heat supplied.

    W=ΔU+Q (16)

Here,

Q is the heat supplied during the compression which is zero for adiabatic compression.

Write the expression for ΔU of both the cylinder and the steel.

  ΔU=nCVΔT+mCΔT (17)

Here,

C is the specific heat capacity of steel in J/kgK.

The value of specific heat capacity is 447J/kgK.

Substitute 0 for Q and the three expressions (8),(9),(16) in (17), then re-arrange to get ΔT,

    W=nCVΔT+mCΔT+0=ΔT(n(52R)+mC)ΔT=W(n(52R)+mC) (18)

Substitute 53.9J for W, 8.314J/molK for R, 9.97×103moles for n, 53.3g for m, 447J/kg-K for C in equation (17) to find ΔU,

    ΔT=53.9J((9.97×103moles)(52×8.314J/molK)+(53.3g×1kg1000g)×(447J/kgK))ΔT=53.9J(0.20722+23.55)=2.242K2.24K

Thus, the increase in temperature of the steel after one compression is 2.24K.

Conclusion:

Therefore, the increase in temperature of the steel after one compression is 2.24K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
No chatgpt pls will upvote
No chatgpt pls will upvote
No chatgpt pls will upvote

Chapter 20 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 20 - In a period of 1.00 s, 5.00 1023 nitrogen...Ch. 20 - A 7.00-L vessel contains 3.50 moles of gas at a...Ch. 20 - Calculate the change in internal energy of 3.00...Ch. 20 - Prob. 10PCh. 20 - In a constant-volume process, 209 J of energy is...Ch. 20 - A vertical cylinder with a heavy piston contains...Ch. 20 - A 1.00-L insulated bottle is full of tea at 90.0C....Ch. 20 - A certain molecule has f degrees of freedom. Show...Ch. 20 - You are working for an automobile tire company....Ch. 20 - Why is the following situation impossible? A team...Ch. 20 - You and your younger brother are designing an air...Ch. 20 - During the compression stroke of a certain...Ch. 20 - Air in a thundercloud expands as it rises. If its...Ch. 20 - Why is the following situation impossible? A new...Ch. 20 - Air (a diatomic ideal gas) at 27.0C and...Ch. 20 - Prob. 22PCh. 20 - Prob. 23PCh. 20 - Prob. 24PCh. 20 - Prob. 25PCh. 20 - The law of atmospheres states that the number...Ch. 20 - Prob. 27APCh. 20 - Prob. 28APCh. 20 - The dimensions of a classroom are 4.20 m 3.00 m ...Ch. 20 - Prob. 30APCh. 20 - The Earths atmosphere consists primarily of oxygen...Ch. 20 - Review. As a sound wave passes through a gas, the...Ch. 20 - Prob. 33APCh. 20 - In a cylinder, a sample of an ideal gas with...Ch. 20 - As a 1.00-mol sample of a monatomic ideal gas...Ch. 20 - A sample consists of an amount n in moles of a...Ch. 20 - The latent heat of vaporization for water at room...Ch. 20 - A vessel contains 1.00 104 oxygen molecules at...Ch. 20 - Prob. 39APCh. 20 - Prob. 40APCh. 20 - Prob. 41APCh. 20 - On the PV diagram for an ideal gas, one isothermal...Ch. 20 - Prob. 43APCh. 20 - Prob. 44APCh. 20 - Prob. 45CP
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Thermodynamics: Crash Course Physics #23; Author: Crash Course;https://www.youtube.com/watch?v=4i1MUWJoI0U;License: Standard YouTube License, CC-BY