Calculate the rms velocity of the molecules of ammonia at S.T.P. Given molecular weight of ammonia `= 17`.

Answer

To calculate the root mean square (RMS) velocity of the molecules of ammonia at standard temperature and pressure (STP), we can follow these steps:

### Step 1: Understand the formula for RMS velocity
The formula for the root mean square velocity (\(V_{rms}\)) of gas molecules is given by:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
or alternatively,
\[ V_{rms} = \sqrt{\frac{3PV}{M}} \]
where:
- \(R\) is the universal gas constant (\(8.314 \, \text{J/(mol K)}\)),
- \(T\) is the temperature in Kelvin,
- \(P\) is the pressure in Pascals,
- \(V\) is the volume in cubic meters,
- \(M\) is the molar mass in kg.

### Step 2: Identify the values at STP
At standard temperature and pressure (STP):
- The standard temperature \(T = 273.15 \, \text{K}\),
- The standard pressure \(P = 1.013 \times 10^5 \, \text{Pa}\),
- The molar mass of ammonia \(M = 17 \, \text{g/mol} = 17 \times 10^{-3} \, \text{kg/mol}\).

### Step 3: Substitute the values into the formula
Using the formula \(V_{rms} = \sqrt{\frac{3PV}{M}}\), we need to calculate the volume \(V\) of one mole of gas at STP, which is \(22.4 \, \text{L} = 22.4 \times 10^{-3} \, \text{m}^3\).

Now substituting the values:
\[ V_{rms} = \sqrt{\frac{3 \times (1.013 \times 10^5) \times (22.4 \times 10^{-3})}{17 \times 10^{-3}}} \]

### Step 4: Calculate the numerator and denominator
First, calculate the numerator:
\[ 3 \times (1.013 \times 10^5) \times (22.4 \times 10^{-3}) = 3 \times 1.013 \times 22.4 \times 10^2 = 68.3 \times 10^2 = 68300 \]

Now, calculate the denominator:
\[ 17 \times 10^{-3} = 0.017 \]

### Step 5: Complete the calculation
Now substitute back into the equation:
\[ V_{rms} = \sqrt{\frac{68300}{0.017}} = \sqrt{4011764.71} \approx 632.8 \, \text{m/s} \]

### Final Answer
The root mean square velocity of the molecules of ammonia at STP is approximately:
\[ V_{rms} \approx 632.8 \, \text{m/s} \]

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