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EXAMPLE 2.14 For any gas, equation 2.48 can be written as (y-1)/y T (A) - 1/1 T₁ For an adiabatic, reversible change in 1 mole of CO2, the pressure changes from 2.44 atm to 0.338 atm. If the initial temperature is 339 K, what is the final tempera- ture? Ignore any contributions from vibrational energy. SOLUTION The major difference between this example and the previous one is that the value of y will be different. For a linear molecule, Cy has contributions from translational ener- gies (Cy(trans) = 3/2R) and rotational energies (Cy(rot) = R). Therefore, we have Cy= 5/2R Because Cp Cy+ R, we also have Gamma, therefore, is Substituting: Y = Solving: C₂ = 7/2R The exponent in the expression is thus Cv 7 Y-1 5 Y = 7 2R 5 2R 7 5 0.338 atm 2/7 2.4 1 75 = 7 Tf 339 K T₁ = 193 K Note how different this answer is from the previous exercise's answer. It can be shown that y - 1 equals for a monatomic ideal gas. Thus, (3) - / T₁

EXAMPLE 2.14 For any gas, equation 2.48 can be written as (y-1)/y T (A) - 1/1 T₁ For an adiabatic, reversible change in 1 mole of CO2, the pressure changes from 2.44 atm to 0.338 atm. If the initial temperature is 339 K, what is the final tempera- ture? Ignore any contributions from vibrational energy. SOLUTION The major difference between this example and the previous one is that the value of y will be different. For a linear molecule, Cy has contributions from translational ener- gies (Cy(trans) = 3/2R) and rotational energies (Cy(rot) = R). Therefore, we have Cy= 5/2R Because Cp Cy+ R, we also have Gamma, therefore, is Substituting: Y = Solving: C₂ = 7/2R The exponent in the expression is thus Cv 7 Y-1 5 Y = 7 2R 5 2R 7 5 0.338 atm 2/7 2.4 1 75 = 7 Tf 339 K T₁ = 193 K Note how different this answer is from the previous exercise's answer. It can be shown that y - 1 equals for a monatomic ideal gas. Thus, (3) - / T₁

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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2.67. Derive the general equation presented in Example 2.14. Start from equation 2.47 and use the combined gas law. EXAMPLE 2.14 For any gas, equation 2.48 can be written as (y-1)/y For an adiabatic, reversible change in 1 mole of CO2, the pressure changes from 2.44 atm to 0.338 atm. If the initial temperature is 339 K, what is the final tempera- ture? Ignore any contributions from vibrational energy. Because Cp Cy + R, we also have SOLUTION The major difference between this example and the previous one is that the value of y will be different. For a linear molecule, Cy has contributions from translational ener- gies (Cy(trans) = 3/2R) and rotational energies (Cy (rot) = R). Therefore, we have Cy= 5/2R Gamma, therefore, is Substituting: The exponent in the expression is thus Solving: C₂ = 7/2R Cp Cy Y-1 Y = 0.338 atm 2.44 atm 7 5 = 7 2R 5 2R I NO 7 5 2/7 Tf T₁ 1 75 7 Tf 339 K T₁ = 193 K Note how different this answer is from the previous exercise's answer. It can be shown that y 1 equals for a monatomic ideal gas. Thus, 2/3 Tf () = 7/1 T₁
Transcribed Image Text:2.67. Derive the general equation presented in Example 2.14. Start from equation 2.47 and use the combined gas law. EXAMPLE 2.14 For any gas, equation 2.48 can be written as (y-1)/y For an adiabatic, reversible change in 1 mole of CO2, the pressure changes from 2.44 atm to 0.338 atm. If the initial temperature is 339 K, what is the final tempera- ture? Ignore any contributions from vibrational energy. Because Cp Cy + R, we also have SOLUTION The major difference between this example and the previous one is that the value of y will be different. For a linear molecule, Cy has contributions from translational ener- gies (Cy(trans) = 3/2R) and rotational energies (Cy (rot) = R). Therefore, we have Cy= 5/2R Gamma, therefore, is Substituting: The exponent in the expression is thus Solving: C₂ = 7/2R Cp Cy Y-1 Y = 0.338 atm 2.44 atm 7 5 = 7 2R 5 2R I NO 7 5 2/7 Tf T₁ 1 75 7 Tf 339 K T₁ = 193 K Note how different this answer is from the previous exercise's answer. It can be shown that y 1 equals for a monatomic ideal gas. Thus, 2/3 Tf () = 7/1 T₁
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