Consider two 1.0-L containers, one containing He(g) at 2.50 atm and 24°C and the other containing Ar(g) at 5.00 atm and 48°C.
-Calculate the ratio of average velocities (He:Ar).

Question

Quizplus · Accepted Answer

The average velocity ($v_{avg}$) of a gas particle can be calculated using the equation derived from the kinetic molecular theory of gases: $v_{avg} = \sqrt{\frac{3RT}{M}}$, where $R$ is the gas constant (8.314 J/mol·K), $T$ is the temperature in Kelvin, and $M$ is the molar mass of the gas in kg/mol.First, convert the temperatures from °C to K:- For He: $T = 24°C + 273.15 = 297.15 K$- For Ar: $T = 48°C + 273.15 = 321.15 K$Molar masses (in kg for the equation's consistency):- He: 4.00 g/mol = 0.004 kg/mol- Ar: 39.95 g/mol = 0.03995 kg/molNow, calculate the average velocities:- For He: $v_{avg,He} = \sqrt{\frac{3 \times 8.314 \times 297.15}{0.004}}$- For Ar: $v_{avg,Ar} = \sqrt{\frac{3 \times 8.314 \times 321.15}{0.03995}}$Calculate the ratio $v_{avg,He} : v_{avg,Ar}$:$$ \frac{v_{avg,He}}{v_{avg,Ar}} = \frac{\sqrt{\frac{3 \times 8.314 \times 297.15}{0.004}}}{\sqrt{\frac{3 \times 8.314 \times 321.15}{0.03995}}} $$Plugging in the numbers:$$ \frac{v_{avg,He}}{v_{avg,Ar}} = \frac{\sqrt{\frac{3 \times 8.314 \times 297.15}{0.004}}}{\sqrt{\frac{3 \times 8.314 \times 321.15}{0.03995}}} \approx 1.41 $$Since the calculated ratio does not match any of the provided options and the calculation involves understanding the relationship between the variables rather than a straightforward computation error, the correct answer is:E) none of these.

Consider Two 1