Question

1. the maxwell - boltzmann distributions of molecular speeds in samples of two different gases at the same temperature are shown. which gas has the greater molar mass?

a) gas a
b) gas b
c) both gases have the same molar mass.
d) it cannot be determined unless the pressure of each sample is known.

1. for parts of the free response question that require calculations, clearly show the method used and the steps involved in arriving at your answers. you must show your work to receive credit for your answer. examples and equations may be included in your answers where appropriate.

dry air is comprised of n₂, o₂, and traces of other gases.
a student obtains a 2.00 l container filled with a sample of dry air at a total pressure of 1.30 atm and a temperature of 295 k.
(a) calculate the total number of moles of gas in the container.
(b) the student cools the gaseous mixture until either n₂ or o₂ condenses. which gas will condense first? justify your answer in terms of the types and relative strengths of the intermolecular forces in n₂ and o₂.
(c) using the legend given, draw at least three molecules of each type of gas in the sealed container represented below when one of the substances has condensed, but the other substance is still a gas.
(d) in this sample, 78% of the molecules are n₂ and 21% of the molecules are o₂. calculate the total number of grams of the substance that condenses first.

Explanation:

Question 6

According to the Maxwell - Boltzmann distribution, at the same temperature, heavier gases have lower most - probable speeds. Gas A has a lower most - probable speed (the peak of the distribution is at a lower molecular speed) compared to gas B. Since molar mass \(M\) and most - probable speed \(v_{p}=\sqrt{\frac{2RT}{M}}\) (where \(R\) is the gas constant and \(T\) is the temperature), a lower \(v_{p}\) implies a higher \(M\).

Step1: Use the ideal gas law

The ideal gas law is \(PV = nRT\), where \(P\) is pressure (\(P = 1.30\ atm\)), \(V\) is volume (\(V=2.00\ L\)), \(n\) is the number of moles (what we want to find), \(R = 0.0821\ L\cdot atm/(mol\cdot K)\) (gas constant), and \(T = 295\ K\).
We can solve for \(n\) by rearranging the formula to \(n=\frac{PV}{RT}\).

Step2: Substitute the values

Substitute \(P = 1.30\ atm\), \(V = 2.00\ L\), \(R=0.0821\ L\cdot atm/(mol\cdot K)\), and \(T = 295\ K\) into the formula:
\(n=\frac{1.30\ atm\times2.00\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times295\ K}\)
\(n=\frac{2.60}{24.2195}\ mol\approx0.107\ mol\)

Both \(N_{2}\) and \(O_{2}\) are non - polar molecules. The only intermolecular forces acting between them are London dispersion forces. The strength of London dispersion forces is related to the molar mass (or the number of electrons). The molar mass of \(O_{2}\) (\(M = 32\ g/mol\)) is greater than that of \(N_{2}\) (\(M = 28\ g/mol\)). A higher molar mass (more electrons) leads to stronger London dispersion forces. Stronger intermolecular forces mean a higher boiling point. So, when cooling, the gas with stronger intermolecular forces (higher boiling point) will condense first.

Answer:

a) Gas A

Question 7(a)