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Chapter 13: Problem 96

If three moles of monatomic gas is mixed with 1 moles diatomic gas, the resultant value of \(\gamma\) for the mixture is (A) \(1.66\) (B) \(1.50\) (C) \(1.40\) (D) \(1.57\)

Short Answer

Expert verified
The effective adiabatic index for the mixture is \(\gamma = \frac{11}{7} \approx 1.57\). Therefore, the correct answer is (D) \(1.57\).

Step by step solution

01

Recall the adiabatic index formula#

In thermodynamics, the adiabatic index (\(\gamma\)) is the ratio of the molar heat capacity at constant pressure (\(C_p\)) to the molar heat capacity at constant volume (\(C_v\)). Mathematically, it is expressed as follows: \[\gamma = \frac{C_p}{C_v}\]
02

Determine the individual heat capacities and adiabatic indexes#

For monatomic and diatomic gases, the molar heat capacities are known: - For monatomic gases, \(C_v = \frac{3}{2}R\) and \(C_p = \frac{5}{2}R\), where R is the gas constant. Therefore, their adiabatic index (let's call it \(\gamma_1\)) is \[\gamma_1 = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}\] - For diatomic gases, \(C_v = \frac{5}{2}R\) and \(C_p = \frac{7}{2}R\). Therefore, their adiabatic index (let's call it \(\gamma_2\)) is \[\gamma_2 = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}\]
03

Calculate the total volumes and heat capacities of the mixture#

First, let's find the total volume of the system (V). Since both gases are ideal, their volumes are additive. With 3 moles of monatomic gas and 1 mole of diatomic gas, the total number of moles of the mixture (n) is \[n = 3 + 1 = 4\] Thus, the volume fraction for monatomic and diatomic gases are \[\frac{V_1}{V} = \frac{3}{4}\] and \[\frac{V_2}{V} = \frac{1}{4}\] Now we can find the total heat capacities of the system at constant volume (Cv) and at constant pressure (Cp) as follows: \[C_v = \frac{3}{4} \cdot \frac{3}{2}R + \frac{1}{4} \cdot \frac{5}{2}R\] \[C_p = \frac{3}{4} \cdot \frac{5}{2}R + \frac{1}{4} \cdot \frac{7}{2}R\]
04

Calculate the effective adiabatic index#

Now that we have the total heat capacities (Cv and Cp), we can calculate the adiabatic index for the mixture using the initial formula: \[\gamma = \frac{C_p}{C_v}\] Substitute the expressions for Cp and Cv as calculated in the previous step: \[\gamma = \frac{\frac{3}{4} \cdot \frac{5}{2}R + \frac{1}{4} \cdot \frac{7}{2}R}{\frac{3}{4} \cdot \frac{3}{2}R + \frac{1}{4} \cdot \frac{5}{2}R}\] Cancel the R terms and simplify: \[\gamma = \frac{15 + 7}{9 + 5} = \frac{22}{14} = \frac{11}{7}\]
05

Determine the answer choice#

Finally, to find the answer, approximate the numeric value of \(\gamma\): \[\gamma = \frac{11}{7} \approx 1.57\] Thus, the correct answer is: (D) \(1.57\).

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