EBK COLLEGE PHYSICS, VOLUME 2
EBK COLLEGE PHYSICS, VOLUME 2
11th Edition
ISBN: 9781337514644
Author: Vuille
Publisher: CENGAGE LEARNING - CONSIGNMENT
Question
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Chapter 10, Problem 6P

(a)

To determine

The value of absolute zero in Celsius.

(a)

Expert Solution
Check Mark

Answer to Problem 6P

The value of absolute zero in Celsius is 270οC .

Explanation of Solution

Section 1:

Explanation:

To determine: The constants a and b.

Answer:

The constant (a) is 4.7×103atm/οC

The constant (b) is 1.27 atm.

Given Info: The pressure ( P1 ) is 0.900 atm when temperature ( T1 ) is 78.5οC . The pressure ( P2 ) is 1.635 atm when temperature ( T2 ) is 78.0οC .

Since, the volume is constant, the P-T graph is a straight line of the form,

P=aT+b                            (I)

  • P is the pressure.
  • T is the temperature.
  • a and b are constants.

For the two sets of pressure and temperature,

P1=aT1+b                           (II)

P2=aT2+b                         (III)

From Equations (I) and (II),

a=P1P2T1T2                         (IV)

Substitute 0.900 atm for P1 , 78.5οC for T1 , 1.635 atm for P2 and 78.0οC for T2 to get a.

a=(0.900atm)(1.635atm)[78.5οC][78.0οC]=4.7×103atm/οC

From Equation (II),

b=P1aT1

Substitute 0.900 atm for P1 , 78.5οC for T1 and 4.7×103atm/οC for a.

b=(0.900atm)[4.7×103atm/οC][78.5οC]=1.27atm

The constant (a) is 4.7×103atm/οC

The constant (b) is 1.27 atm.

Conclusion:

From Equation (I), the temperature is,

T=bPa

At absolute zero, pressure is zero.

Substitute 0 atm for P, 4.7×103atm/οC for a and 1.27 atm for b to get T.

T=(0.atm)(1.27atm)4.7×103atm/οC=270.21οC

On Rounding off,

T=270οC

(b)

To determine

The pressure.

(b)

Expert Solution
Check Mark

Answer to Problem 6P

The pressure is 1.27 atm.

Explanation of Solution

From Equation (I) of (a),

P=aT+b

At the freezing point, temperature is zero.

Substitute 0οC for T, 4.7×103atm/οC for a and 1.27 atm for b to get P.

P=(4.7×103atm/οC)(0οC)+(1.27atm)=1.27atm

Conclusion:

The pressure is 1.27 atm.

(c)

To determine

The pressure.

(c)

Expert Solution
Check Mark

Answer to Problem 6P

The pressure is 1.74 atm.

Explanation of Solution

From Equation (I) of (a),

P=aT+b

At the boiling point, temperature is 100οC .

Substitute 100οC for T, 4.7×103atm/οC for a and 1.27 atm for b to get P.

P=(4.7×103atm/οC)(100οC)+(1.27atm)=1.74atm

Conclusion:

The pressure is 1.74 atm.

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Chapter 10 Solutions

EBK COLLEGE PHYSICS, VOLUME 2

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