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Chapter 18: Problem 58

A gas mixture contains monatomic argon and diatomic oxygen. An adiabatic expansion that doubles its volume results in the pressure dropping to one- third of its original value. What fraction of the molecules are argon?

Short Answer

Expert verified
The fraction of the molecules in the mixture that are argon is approximately 0.44, or 44%.

Step by step solution

01

Understand the problem and gather information

From the problem, the following information is relevant: after adiabatic expansion, volume doubles (i.e., final volume \(V_f = 2V_i\) where \(V_i\) is initial volume) and pressure drops to one-third of its original value (i.e., final pressure \(P_f = P_i/3\) where \(P_i\) is initial pressure). We want to find what fraction of the molecules are argon.
02

Understand Adiabatic Process for different kinds of gases

An adiabatic process is one in which no heat is gained or lost. The process is governed by \(P_i V_i^{\gamma_i} = P_f V_f^{\gamma_i}\) where \(\gamma_i\) is the adiabatic constant of the relevant gas. For a monoatomic ideal gas like argon, \(\gamma_i = 5/3\); for a diatomic ideal gas like oxygen, \(\gamma_i = 7/5\).
03

Apply Adiabatic Process formula for argon

Let \(N_a\) be the number of argon molecules, \(N_o\) the number of oxygen molecules, \(P_a\) argon's pressure, and \(P_o\) oxygen's pressure. Applying the adiabatic process formula for argon, we have \(P_a V_i^{5/3} = (P_a/3) (2V_i)^{5/3}\). This simplifies to \(P_a = 3/2^{5/3} P_a\).
04

Apply Adiabatic Process formula for oxygen

For oxygen, the formula gives \(P_o V_i^{7/5} = (P_o/3) (2V_i)^{7/5}\). Solving this gives \(P_o = 3/2^{7/5} P_o\).
05

Apply Ideal Gas Law

The ideal gas law is \(PV = nRT\), where \(n\) is the number of moles \(N\), \(R\) is the gas constant, and \(T\) is the absolute temperature. If we apply this for argon and oxygen, we have \(P_a = N_a RT/V_i\) and \(P_o = N_o RT/V_i\).
06

Combine the equations

Substituting the expressions for \(P_a\) and \(P_o\) from Step 5 into the results from Steps 3 and 4 gives \(N_a/N = 3/2^{5/3}\) and \(N_o/N = 3/2^{7/5}\), where \(N = N_a + N_o\), the total number of molecules.
07

Calculate the fraction of argon molecules

The fraction of argon molecules is \(N_a/N\). Adding the two fractions from Step 6 and equating to 1 (since \(N_a/N + N_o/N = 1\)), we can solve for \(N_a/N\). This gives \(N_a/N = 3/2^{5/3} /(3/2^{5/3} + 3/2^{7/5}) \approx 0.44\).

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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.

Prove that the slope of an adiabat at a given point in a \(p V\) diagram is \(\gamma\) times the slope of the isotherm passing through the same point.

A gasoline engine has compression ratio 8.5 (see Example 18.3 for the meaning of this term), and the fuel-air mixture compresses adiabatically with \(\gamma=1.4 .\) If the mixture enters the engine at \(30^{\circ} \mathrm{C},\) what will its temperature be at maximum compression?

Experimental studies show that the \(p V\) curve for a frog's lung can be approximated by \(p=10 v^{3}-67 v^{2}+220 v,\) with \(v\) in \(\mathrm{mL}\) and \(p\) in \(\mathrm{Pa}\). Find the work done when such a lung inflates from zero to \(4.5 \mathrm{mL}\) volume.

In a closed but uninsulated container, \(500 \mathrm{g}\) of water are shaken violently until the temperature rises by \(3.0^{\circ} \mathrm{C}\). The mechanical work required in the process is \(9.0 \mathrm{kJ}\). (a) How much heat is transferred during the shaking? (b) How much mechanical energy would have been required if the container had been perfectly insulated?
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