To calculate the root mean square speed (\(v_{\text{rms}}\)) of ammonia (NH3) molecules at a temperature of 50 degrees Celsius, we utilize the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
In this equation, \(R\) is the ideal gas constant, \(T\) is the absolute temperature in Kelvin, and \(M\) is the molar mass of the gas in kilograms per mole.
First, we convert the temperature from degrees Celsius to Kelvin:
\[ T = 50 + 273.15 = 323.15 \, \text{K} \]
Next, we identify the appropriate value for \(R\). Since we are working with speed, we use:
\[ R = 8.314 \, \text{J/(mol·K)} \]
We also need the molar mass of ammonia. The molar mass of NH3 is calculated as follows:
- Nitrogen (N): 14.007 g/mol
- Hydrogen (H): 1.008 g/mol × 3 = 3.024 g/mol
- Total: 14.007 + 3.024 = 17.031 g/mol
To convert this to kilograms per mole, we divide by 1000:
\[ M = \frac{17.031 \, \text{g/mol}}{1000} = 0.017031 \, \text{kg/mol} \]
Now, substituting the values into the \(v_{\text{rms}}\) formula:
\[ v_{\text{rms}} = \sqrt{\frac{3 \times 8.314 \, \text{J/(mol·K)} \times 323.15 \, \text{K}}{0.017031 \, \text{kg/mol}}} \]
Next, we simplify the units. The joules can be expressed in terms of base units:
\[ \text{J} = \text{kg} \cdot \text{m}^2/\text{s}^2 \]
Thus, when substituting, the units of kilograms cancel out, moles cancel out, and kelvins cancel out, leaving us with:
\[ \text{m}^2/\text{s}^2 \]
Taking the square root of this gives us the final unit of:
\[ \text{m/s} \]
After performing the calculation, we find:
\[ v_{\text{rms}} \approx 687.9 \, \text{m/s} \]
It is important to note that while the temperature was given with one significant figure, for accuracy in calculations, we used four significant figures in our final answer. Rounding to one significant figure would yield 700 m/s, which is less precise than the calculated value.
In summary, understanding the relationships between temperature, molar mass, and the ideal gas constant is crucial for accurately determining the root mean square speed of gas molecules.
