Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 20, Problem 15P

(a)

To determine

The final temperature.

(a)

Expert Solution
Check Mark

Answer to Problem 15P

The final temperature is 380 K .

Explanation of Solution

The heat lost from the oxygen at higher temperature must be equal to the heat gained by oxygen at lower temperature.

Write the equation for the heat change of the system.

  Qcold=Qhot                                                                                                         (I)

Here, Qcold is the heat gained by oxygen at lower temperature and Qhot is the heat lost from oxygen at higher temperature.

Write the equation for Qcold .

  Qcold=mcoldc(TT1)                                                                                          (II)

Here, mcold is the mass of the oxygen at lower temperature, c is the specific heat capacity of oxygen, T is the final temperature and T1 is the initial temperature of the oxygen at lower temperature.

The mass can be expressed as the product of molar mass and the number of moles of the gas.

Write the expression for mcold .

  mcold=n1M

Here, n1 is the number of moles of oxygen at lower temperature and M is the molar mass of oxygen.

Put the above equation in equation (II).

  Qcold=n1Mc(TT1)                                                                                         (III)

Write the equation for Qhot .

  Qhot=mhotc(TT2)                                                                                             (IV)

Here, mhot is the mass of oxygen at higher temperature and T2 is the initial temperature of oxygen at higher temperature.

Write the expression for mhot .

  mhot=n2M

Here, n2 is the number of moles of oxygen at higher temperature.

Put the above equation in equation (IV).

  Qhot=n2Mc(TT2)                                                                                             (V)

Put equations (III) and (V) in equation (I).

  n1Mc(TT1)=n2Mc(TT2)n1(TT1)=n2(TT2)                                                                              (VI)

Write the ideal gas equation.

  PV=nRT

Here, P is the pressure of the gas, V is the volume of the gas, R is the universal gas constant and T is the temperature of the gas.

Rewrite the above equation for n .

  n=PVRT                                                                                                               (VII)

Use equation (VII) to write the expression for n1 .

  n1=P1V1RT1                                                                                                              (VIII)

Here, P1 is the pressure of oxygen at lower temperature and V1 is the volume of oxygen at lower temperature.

Use equation (VII) to write the expression for n2 .

  n2=P2V2RT2                                                                                                                (IX)

Here, P2 is the pressure of oxygen at higher temperature and V2 is the volume of oxygen at higher temperature.

Put equations (VIII) and (IX) in equation (VI) and rewrite it for T .

  P1V1RT1(TT1)=P2V2RT2(TT2)P1V1T1TP1V1T1T1=P2V2T2T+P2V2T2T2T(P1V1T1+P2V2T2)=P1V1+P2V2T=P1V1+P2V2(P1V1T1+P2V2T2)                                                                   (X)

Conclusion:

Substitute 1.75 atm for P1 , 16.8 L for V1 , 300 K for T1 , 2.25 atm for P2 , 22.4 L for V2 and 450 K for T2 in equation (X) to find T .

  T=(1.75 atm)(16.8 L)+(2.25 atm)(22.4 L)((1.75 atm)(16.8 L)300 K+(2.25 atm)(22.4 L)450 K)=380 K

Therefore, the final temperature is 380 K .

(b)

To determine

The final pressure.

(b)

Expert Solution
Check Mark

Answer to Problem 15P

The final pressure is 2.04 atm .

Explanation of Solution

Rewrite the ideal gas equation for pressure.

  P=nRTV

Use the above equation to write the expression for the final pressure of the oxygen.

  Pf=nfRTVf                                                                                                             (XI)

Here, nf is the number of moles in the final state, Pf is the final pressure and Vf is the final volume.

For the final state of the system, the value of n will be equal to the sum of n1 and n2 , the value of V will be equal to the sum of V1 and V2 .

Write the equation for nf .

  nf=n1+n2

Put equations (VIII) and (IX) in the above equation.

  nf=P1V1RT1+P2V2RT2=1R(P1V1T1+P2V2T2)                                                                                            (XII)

Write the equation for Vf .

  Vf=V1+V2                                                                                                        (XIII)

Put equations (X), (XII) and (XIII) in equation (XI).

  Pf=(1R(P1V1T1+P2V2T2))RV1+V2P1V1+P2V2(P1V1T1+P2V2T2)=P1V1+P2V2V1+V2                                                    (XIV)

Conclusion:

Substitute 1.75 atm for P1 , 16.8 L for V1 , 2.25 atm for P2 and 22.4 L for V2 in equation (XIV) to find Pf .

  Pf=(1.75 atm)(16.8 L)+(2.25 atm)(22.4 L)16.8 L+22.4 L=2.04 atm

Therefore, the final pressure is 2.04 atm .

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Chapter 20 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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