Question

arrange the following molecules in order of increasing average molecular speed. ar at 396k, ar at 345k, o₂ at 396k, nh₃ at 454k enter formulas and temperatures in the boxes below: 1 = slowest, 4 = fastest 1 at k 2 at k 3 at k 4 at k

Explanation:

Step1: Recall the formula for average molecular speed

The formula for the average molecular speed \(v_{avg}=\sqrt{\frac{8RT}{\pi M}}\), where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas. From this formula, we can see that \(v_{avg}\propto\sqrt{\frac{T}{M}}\).

Step2: Calculate the \(\frac{T}{M}\) ratio for each gas

• For \(Ar\) at \(T = 345K\), \(M(Ar)=39.948g/mol\), \(\frac{T}{M}=\frac{345}{39.948}\approx8.64\)
• For \(Ar\) at \(T = 396K\), \(M(Ar) = 39.948g/mol\), \(\frac{T}{M}=\frac{396}{39.948}\approx9.91\)
• For \(O_{2}\) at \(T = 396K\), \(M(O_{2})=32.00g/mol\), \(\frac{T}{M}=\frac{396}{32.00}=12.375\)
• For \(NH_{3}\) at \(T = 454K\), \(M(NH_{3}) = 17.03g/mol\), \(\frac{T}{M}=\frac{454}{17.03}\approx26.66\)

Step3: Compare the \(\frac{T}{M}\) ratios

Since \(v_{avg}\propto\sqrt{\frac{T}{M}}\), the order of \(\frac{T}{M}\) ratios (from smallest to largest) will give the order of average molecular speeds (from slowest to fastest). The order of \(\frac{T}{M}\) is: \(8.64(Ar - 345K)<9.91(Ar - 396K)<12.375(O_{2}-396K)<26.66(NH_{3}-454K)\)

Answer:

1. \(Ar\) at \(345\) K
2. \(Ar\) at \(396\) K
3. \(O_{2}\) at \(396\) K
4. \(NH_{3}\) at \(454\) K