Question

question 3
calculate the root mean square speed of oxygen gas molecules at 50°c degrees celsius. m/s
type numbers in the boxes.
8 points
question 4
samples of ar gas and kr gas are both at 300 k.
the average kinetic energy of the ar is the average kinetic energy of the kr.
the average speed of the ar is the average speed of the kr.
the rate of effusion of the ar is the rate of effusion of the kr.
fill in the blanks by selecting one option from each menu.
part 1: 3 points
part 2: 3 points
part 3: 3 points
9 points

Explanation:

Question 3

Step1: Convert temperature to Kelvin

The formula for root - mean - square speed \(v_{rms}=\sqrt{\frac{3RT}{M}}\), where \(R = 8.314\ J/(mol\cdot K)\), \(T\) is in Kelvin, and \(M\) is the molar mass in \(kg/mol\).
First, convert \(T = 50^{\circ}C\) to Kelvin: \(T=(50 + 273.15)\ K=323.15\ K\).
The molar mass of \(O_2\) is \(M = 32\times10^{- 3}\ kg/mol\)

Step2: Calculate \(v_{rms}\)

Substitute the values into the formula \(v_{rms}=\sqrt{\frac{3RT}{M}}\)

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$$ \frac{8060.0173}{0.032}=251875.5406 $$

\(v_{rms}=\sqrt{251875.5406}\approx502\ m/s\)

• Average kinetic energy: The average kinetic energy of gas molecules is given by \(\overline{KE}=\frac{3}{2}kT\) (where \(k\) is the Boltzmann constant and \(T\) is the temperature). Since \(T = 300\ K\) for both \(Ar\) and \(Kr\), the average kinetic energy of \(Ar\) is equal to the average kinetic energy of \(Kr\) (\(\overline{KE}\propto T\), and \(T\) is the same).
• Average speed: The average speed of gas molecules is \(v_{avg}=\sqrt{\frac{8RT}{\pi M}}\). The molar mass of \(Ar(M_{Ar}=40\ g/mol)\) and \(Kr(M_{Kr} = 83.8\ g/mol)\). Since \(v_{avg}\propto\frac{1}{\sqrt{M}}\) (when \(T\) is constant), and \(M_{Ar}
• Rate of effusion: Graham's law of effusion states that \(r\propto\frac{1}{\sqrt{M}}\) (where \(r\) is the rate of effusion). Since \(M_{Ar}

Answer:

\(502\)

Question 4