Exercises

Exercise 1. Calculate the rms speed of N₂ at 98.2°C.

The molar mass of N₂ = 28.0 g/mol = 0.028 kg/mol
T = 98.2°C + 273.15 = 371.35 K
R = 8.314 \(\frac{kg⋅m^2}{s^2⋅mol⋅K}\)

\(\displaystyle \sqrt{\frac{3\times\;8.314\frac{kg⋅m^2}{s^2⋅mol⋅K}\times\;371.35\;K}{0.028\frac{kg}{mol}}}\;=\;\mathbf{5.75\times 10^2\;\frac{m}{s}}\)

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Exercise 2. Uranium hexafluride, UF₆, sublimes at 57°C under normal atmospheric pressure. What is the rms speed of UF₆ at 57°C? Report your answer in m/s.

The molar mass of UF₆ = 352.02 g/mol = 0.35202 kg/mol
T = 57°C + 273.15 = 330 K
R = 8.314 \(\frac{kg⋅m^2}{s^2⋅mol⋅K}\)

\(\displaystyle \sqrt{\frac{3\times\;8.314\frac{kg⋅m^2}{s^2⋅mol⋅K}\times\;330\;K}{0.35202\frac{kg}{mol}}}\;=\;\mathbf{1.53\times 10^2\;\frac{m}{s}}\)

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Exercise 3. What is the molar mass of a gas that diffuses 1.98 times faster than Xe gas.

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;of\;gas\;A}}{\mathrm{rate\;of\;effusion\;of\;Xe}}\;=\;\sqrt{\frac{M_mXe}{M_mA}}\)

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;of\;Gas\;A}}{\mathrm{rate\;of\;effusion\;of\;Xe}}\;=\;1.98\;=\sqrt{\frac{131.29\frac{g}{mol}}{M_m\;Gas\;A}}\)

Square both sides of the equation and solve for M_m of Gas A.

\(\displaystyle (1.98)^2\;=\frac{131.29\frac{g}{mol}}{M_m\;Gas\;A}\)
 
\(\displaystyle M_m\;Gas\;A\;=\;\frac{131.29\frac{g}{mol}}{1.98^2}\;=\;\mathbf{33.49\frac{g}{mol}}\)

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Exercise 4. What is the molar mass of a gas that diffuses 1.87 times slower than Kr gas?

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;of\;gas\;A}}{\mathrm{rate\;of\;effusion\;of\;Kr}}\;=\;\sqrt{\frac{M_mKr}{M_mA}}\)

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;of\;Gas\;A}}{\mathrm{rate\;of\;effusion\;of\;Kr}}\;=\;\frac{1}{1.87}\;=\sqrt{\frac{83.798\frac{g}{mol}}{M_m\;Gas\;A}}\)

Square both sides of the equation and solve for M_m of Gas A.

\(\displaystyle \biggl(\frac{1}{1.87}\biggr)^2\;=\frac{83.798\frac{g}{mol}}{M_m\;Gas\;A}\)
 
\(\displaystyle M_m\;Gas\;A\;=\;\frac{83.798\frac{g}{mol}}{\frac {1}{1.87^2}}\;=\;\mathbf{293\frac{g}{mol}}\)

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Exercise 5. Calculate the ratio of the rates of effusion of N₂ and N₂O under the same conditions.

The molar mass of N₂O is 43.999 g/mol and N₂ is 28.0 g/mol.

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;of\;N_2}}{\mathrm{rate\;of\;effusion\;of\;N_2O}}\;=\;\sqrt{\frac{43.999\frac{g}{mol}}{28.0\frac{g}{mol}}}\;=\;\mathbf{1.25}\)

The N₂ effuses 1.25 times faster than N₂O.

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Exercise 6. A given volume of N₂ takes 64.6 s to effuse. Another gas, under the same conditions, took 88.2 s to effuse. Calculate the molar mass of this gas.

\(\displaystyle \frac{t_{gas}}{t_{N_2}}\;=\;\sqrt{\frac{M_m\;gas}{M_m\;N_2}}\)
 
\(\displaystyle \frac{88.2\;s}{64.6\;s}\;=\;\sqrt{\frac{M_m\;gas}{28.0\frac{g}{mol}}}\)

Square both sides of the equation and solve for the molar mass of the gas.

\(\displaystyle \biggl(\frac{88.2\;s}{64.6\;s}\biggr)^2\;=\;\frac{M_m\;gas}{28.0\frac{g}{mol}}\)
 
M_m gas = 52.2 g/mol

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