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Chapter 18: Problem 59

How much of a triatomic gas with \(C_{V}=3 R\) would you have to add to 10 mol of monatomic gas to get a mixture whose thermodynamic behavior was like that of a diatomic gas?

Short Answer

Expert verified
Isolate \(n_{2}\) in the equation and solve, giving us the amount of the triatomic gas to be added.

Step by step solution

01

List known values

The molecular heat capacity at constant volume for monatomic gas is \(C_{V} = R\), for diatomic gas is \(C_{V} = 2R\), and for the unspecified triatomic gas in the question, it is given as \(C_{V} = 3R\). We have 10 mol of the monatomic gas. Now, we want to find out how much more gas, \(n\), is needed.
02

Set up the equation based on the thermodynamic behavior

The effective molecular heat capacity (\(C_{V}\)) of a mixture is given by the sum of the product of the number of moles (\(n\)) and the molar heat capacity (\(C_{V}\)) for each component, divided by the total moles. So here we have the equation:\[C_{V,eq} = \frac{n_{1}C_{V,1}+n_{2}C_{V,2}}{n_{1}+n_{2}}\]Given \(n_{1}=10\), \(C_{V,1}=R\) (for monatomic gas), \(C_{V,2}=3R\) (for triatomic gas), and we want \(C_{V,eq} = 2R\) (thermodynamic behavior like diatomic gas). We have to find \(n_{2}\), the amount of the triatomic gas to be added.
03

Solve for the unknown

Substitute the known values into the equation and solve for \(n_{2}\). \[2R = \frac{10R + n_{2}3R}{10 + n_{2}}\]By solving this equation, we find the value of \(n_{2}\) (amount of the triatomic gas to be added).

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