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Chapter 18: Problem 28

What should be the approximate specific-heat ratio of a gas consisting of \(50 \% \mathrm{NO}_{2}(\gamma=1.29), 30 \% \mathrm{O}_{2}(\gamma=1.40),\) and \(20 \%\) \(\operatorname{Ar}(\gamma=1.67) ?\)

Short Answer

Expert verified
The approximate specific-heat ratio of the gas mixture is \(1.399\).

Step by step solution

01

Identify the Proportions and Specific Heat Ratios

Identify the proportion (percentage) of each gas in the mixture, as well as its specific heat ratio γ. We have \(50\% NO_{2}\) with \(\gamma=1.29\), \(30\% O_{2}\) with \(\gamma=1.40\) and \(20\% Ar\) with \(\gamma=1.67\)
02

Calculate the Weighted Average

Calculate the weighted average of the specific heat ratios. This is done by multiplying the specific heat ratio of each gas by its proportion in the mixture and then summing these values. The formula is \(\gamma_{avg} = (p_1*\gamma_1)+(p_2*\gamma_2)+..+(p_n*\gamma_n)\). Using the provided proportions and specific heat ratios, we have : \(\gamma_{avg} = 0.5*1.29 + 0.3*1.40 + 0.2*1.67\)
03

Simplify the expression and get the final answer

Simplify the expression obtained in Step 2, and this will yield the specific heat ratio for the entire gas mixture. So \(\gamma_{avg} = 0.645 + 0.42 + 0.334 = 1.399\)

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Most popular questions from this chapter

The adiabatic lapse rate is the rate at which air cools as it rises and expands adiabatically in the atmosphere (see Application: Smog Alert, on page 302 ). Express \(d T\) in terms of \(d p\) for an adiabatic process, and use the hydrostatic equation (Equation 15.2 ) to express \(d p\) in terms of \(d y .\) Then, calculate the lapse rate \(d T / d y .\) Take air's average molecular weight to be 29 u and \(\gamma=1.4,\) and remember that the altitude \(y\) is the negative of the depth \(h\) in Equation 15.2.

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