Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 1, Problem 1C.8BE

(i)

Interpretation Introduction

Interpretation: The temperature and pressure at which 1.0mol of H2O will be in states that corresponds to 1.0mol of N2 at 1.0atm and 25°C has to be calculated.

Concept introduction: According to the principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.

(i)

Expert Solution
Check Mark

Answer to Problem 1C.8BE

The temperature and pressure at which 1.0mol of H2O corresponds to the state of 1.0mol N2 at 1.0atm and 25°C is 5754.738K_ and 1.72atm_, respectively.

Explanation of Solution

The pressure of N2 is 1atm and temperature of N2 is 25°C.

The conversion of given temperature (°C) into K is shown below.

  K=°C+273=25°C+273=298K

The reduced pressure and temperature is given by,

    pt=ppc                                                                                                         (1)

  Tt=TTc                                                                                                           (2)

Where,

  • Ø  Tc is the critical temperature.
  • Ø  pc is the critical pressure.
  • Ø  p is the pressure.
  • Ø  T is the temperature.

The critical temperature of N2 is 33.54K and the critical pressure of N2 is 126.3atm.

Substitute the critical pressure of N2 in equation (1).

    pt=1.0atm126.3atm=0.0079

Hence, the reduced pressure of N2 is 0.0079_.

Substitute the critical temperature of N2 in equation (2).

    Tt=298.15K33.54K=8.889

Hence, the reduced temperature of N2 is 8.889_.

The critical temperature of H2O is 647.4K and the critical pressure of H2O is 218.3atm.

According to the principle of corresponding states, the value of reduced pressure for H2O is same as that of N2.  Thus, the reduced pressure for H2O is 0.0079.

The pressure at which 1.0mol of H2O is in corresponding the state of N2 can be calculated using the equation shown below.

    p=pt×pc                                                                                                     (3)

Substitute the reduced pressure and critical pressure of H2O in equation (3).

    p=0.0079×218.3atm=1.72atm_

Hence, the pressure of 1.0mol of H2O is 1.72atm_.

The temperature at which 1.0mol of H2O is in corresponding the state of H2 can be calculated using the equation shown below,

    T=Tt×Tc                                                                                                      (4)

Substitute the reduced temperature and critical temperature of H2O in equation (4).

    T=8.889×647.4K=5754.738K_

Hence, the temperature of 1.0mol of H2O is 5754.738K_.

(ii)

Interpretation Introduction

Interpretation: The temperature and pressure at which 1.0mol of CO2 will be in states that corresponds to 1.0mol of N2 at 1.0atm and 25°C.

Concept introduction: According to the principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.

(ii)

Expert Solution
Check Mark

Answer to Problem 1C.8BE

The temperature and pressure at which 1.0mol of CO2 is in corresponding the state of N2 is 2704.0338K_ and 0.57591atm_, respectively.

Explanation of Solution

The pressure of N2 is 1atm and temperature of N2 is 25°C.

The conversion of given temperature (°C) into K is shown below.

K=°C+273=25°C+273=298K

The reduced pressure and temperature is given by,

    pt=ppc (1)

  Tt=TTc                                                                                                           (2)

Where,

  • Ø  Tc is the critical temperature.
  • Ø  pc is the critical pressure.
  • Ø  p is the pressure.
  • Ø  T is the temperature.

The critical temperature of N2 is 33.54K and the critical pressure of N2 is 126.3atm.

Substitute the critical pressure of N2 in equation (1).

    pt=1.0atm126.3atm=0.0079

Hence, the reduced pressure of N2 is 0.0079_.

Substitute the critical temperature of N2 in equation (2).

    Tt=298.15K33.54K=8.889

Hence, the reduced temperature of N2 is 8.889_.

The critical temperature of CO2 is 304.2K and the critical pressure of CO2 is 72.9atm.

According to the principle of corresponding states, the value of reduced pressure for CO2 is same as that of N2.  Thus, the reduced pressure for CO2 is 0.0079.

The pressure at which 1.0mol of CO2 is in corresponding the state of N2 can be calculated using the equation shown below.

    p=pt×pc                                                                                                     (3)

Substitute the reduced pressure and critical pressure of CO2 in equation (3).

    p=0.0079×72.9atm=0.57591atm_

Hence, the pressure of 1.0mol of CO2 is 0.57591atm_.

The temperature at which 1.0mol of CO2 is in corresponding the state of N2 can be calculated using the equation shown below,

    T=Tt×Tc                                                                                                      (4)

Substitute the reduced temperature and critical temperature of CO2 in equation (4).

    T=8.889×304.2K=2704.0338K_

Hence, the temperature of 1.0mol of CO2 is 2704.0338K_.

(iii)

Interpretation Introduction

Interpretation: The temperature and pressure at which 1.0mol of Ar will be in states that corresponds to 1.0mol of N2 at 1.0atm and 25°C.

Concept introduction: According to the principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.

(iii)

Expert Solution
Check Mark

Answer to Problem 1C.8BE

The temperature at which 1.0mol of Ar is in corresponding the state of N2 is 1339.57K_ and 0.3792atm_, respectively.

Explanation of Solution

The pressure of N2 is 1atm and temperature of N2 is 25°C.

The conversion of given temperature (°C) into K is shown below.

  K=°C+273=25°C+273=298K

The principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.  The reduced pressure and temperature is given by,

    pt=ppc  (1)

  Tt=TTc     (2)

Where,

  • Ø  Tc is the critical temperature.
  • Ø  pc is the critical pressure.
  • Ø  p is the pressure.
  • Ø  T is the temperature.

The critical temperature of N2 is 33.54K and the critical pressure of N2 is 126.3atm.

Substitute the critical pressure of N2 in equation (1).

    pt=1.0atm126.3atm=0.0079

Hence, the reduced pressure of N2 is 0.0079_.

Substitute the critical temperature of N2 in equation (2).

    Tt=298.15K33.54K=8.889

Hence, the reduced temperature of N2 is 8.889_.

The critical temperature of Ar is 150.7K and the critical pressure of Ar is 48atm.

According to the principle of corresponding states, the value of reduced pressure for Ar is same as that of N2.  Thus, the reduced pressure for Ar is 0.0079.

The pressure at which 1.0mol of Ar is in corresponding the state of N2 can be calculated using the equation shown below.

    p=pt×pc                                                                                                    (3)

Substitute the reduced pressure and critical pressure of Ar in equation (3).

    p=0.0079×48atm=0.3792atm_

Hence, the pressure of 1.0mol of Ar is 0.3792atm_.

The temperature at which 1.0mol of Ar is in corresponding the state of N2 can be calculated using the equation shown below,

    T=Tt×Tc                                                                                                     (4)

Substitute the reduced temperature and critical temperature of Ar in equation (4).

    T=8.889×150.7K=1339.57K_

Hence, the temperature of 1.0mol of Ar is 1339.57K_.

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Chapter 1 Solutions

Atkins' Physical Chemistry

Ch. 1 - Prob. 1A.3BECh. 1 - Prob. 1A.4AECh. 1 - Prob. 1A.4BECh. 1 - Prob. 1A.5AECh. 1 - Prob. 1A.5BECh. 1 - Prob. 1A.6AECh. 1 - Prob. 1A.6BECh. 1 - Prob. 1A.7AECh. 1 - Prob. 1A.7BECh. 1 - Prob. 1A.8AECh. 1 - Prob. 1A.8BECh. 1 - Prob. 1A.9AECh. 1 - Prob. 1A.9BECh. 1 - Prob. 1A.10AECh. 1 - Prob. 1A.10BECh. 1 - Prob. 1A.11AECh. 1 - Prob. 1A.11BECh. 1 - Prob. 1A.1PCh. 1 - Prob. 1A.2PCh. 1 - Prob. 1A.3PCh. 1 - Prob. 1A.4PCh. 1 - Prob. 1A.5PCh. 1 - Prob. 1A.6PCh. 1 - Prob. 1A.7PCh. 1 - Prob. 1A.8PCh. 1 - Prob. 1A.9PCh. 1 - Prob. 1A.10PCh. 1 - Prob. 1A.11PCh. 1 - Prob. 1A.12PCh. 1 - Prob. 1A.13PCh. 1 - Prob. 1A.14PCh. 1 - Prob. 1B.1DQCh. 1 - Prob. 1B.2DQCh. 1 - Prob. 1B.3DQCh. 1 - Prob. 1B.1AECh. 1 - Prob. 1B.1BECh. 1 - Prob. 1B.2AECh. 1 - Prob. 1B.2BECh. 1 - Prob. 1B.3AECh. 1 - Prob. 1B.3BECh. 1 - Prob. 1B.4AECh. 1 - Prob. 1B.4BECh. 1 - Prob. 1B.5AECh. 1 - Prob. 1B.5BECh. 1 - Prob. 1B.6AECh. 1 - Prob. 1B.6BECh. 1 - Prob. 1B.7AECh. 1 - Prob. 1B.7BECh. 1 - Prob. 1B.8AECh. 1 - Prob. 1B.8BECh. 1 - Prob. 1B.9AECh. 1 - Prob. 1B.9BECh. 1 - Prob. 1B.1PCh. 1 - Prob. 1B.2PCh. 1 - Prob. 1B.3PCh. 1 - Prob. 1B.4PCh. 1 - Prob. 1B.5PCh. 1 - Prob. 1B.6PCh. 1 - Prob. 1B.7PCh. 1 - Prob. 1B.8PCh. 1 - Prob. 1B.9PCh. 1 - Prob. 1B.10PCh. 1 - Prob. 1B.11PCh. 1 - Prob. 1C.1DQCh. 1 - Prob. 1C.2DQCh. 1 - Prob. 1C.3DQCh. 1 - Prob. 1C.4DQCh. 1 - Prob. 1C.1AECh. 1 - Prob. 1C.1BECh. 1 - Prob. 1C.2AECh. 1 - Prob. 1C.2BECh. 1 - Prob. 1C.3AECh. 1 - Prob. 1C.3BECh. 1 - Prob. 1C.4AECh. 1 - Prob. 1C.4BECh. 1 - Prob. 1C.5AECh. 1 - Prob. 1C.5BECh. 1 - Prob. 1C.6AECh. 1 - Prob. 1C.6BECh. 1 - Prob. 1C.7AECh. 1 - Prob. 1C.7BECh. 1 - Prob. 1C.8AECh. 1 - Prob. 1C.8BECh. 1 - Prob. 1C.9AECh. 1 - Prob. 1C.9BECh. 1 - Prob. 1C.1PCh. 1 - Prob. 1C.2PCh. 1 - Prob. 1C.3PCh. 1 - Prob. 1C.4PCh. 1 - Prob. 1C.5PCh. 1 - Prob. 1C.6PCh. 1 - Prob. 1C.7PCh. 1 - Prob. 1C.8PCh. 1 - Prob. 1C.9PCh. 1 - Prob. 1C.10PCh. 1 - Prob. 1C.11PCh. 1 - Prob. 1C.12PCh. 1 - Prob. 1C.13PCh. 1 - Prob. 1C.14PCh. 1 - Prob. 1C.15PCh. 1 - Prob. 1C.16PCh. 1 - Prob. 1C.17PCh. 1 - Prob. 1C.18PCh. 1 - Prob. 1C.19PCh. 1 - Prob. 1C.20PCh. 1 - Prob. 1C.22PCh. 1 - Prob. 1C.23PCh. 1 - Prob. 1C.24PCh. 1 - Prob. 1.1IACh. 1 - Prob. 1.2IACh. 1 - Prob. 1.3IA
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