Find the average of magnitude of linear momentum of helium molecules in a sample of helium gas at temperature of `150 piK`. Mass of a helium molecules `=(166//3) xx 10^(-27) kg` and `R = 25//3J-mol^(-1)K^(-1)`

Answer

To find the average magnitude of linear momentum of helium molecules in a sample of helium gas at a temperature of \(150 \pi K\), we will follow these steps:

### Step 1: Understand the formula for average linear momentum
The average linear momentum \(p_{\text{average}}\) of a molecule is given by:
\[ p_{\text{average}} = m \cdot v_{\text{average}} \]
where \(m\) is the mass of a helium molecule and \(v_{\text{average}}\) is the average velocity of the molecule.

### Step 2: Find the formula for average velocity
The average velocity \(v_{\text{average}}\) of gas molecules can be calculated using the formula:
\[ v_{\text{average}} = \sqrt{\frac{8RT}{\pi m}} \]
where:
- \(R\) is the universal gas constant,
- \(T\) is the temperature in Kelvin,
- \(m\) is the mass of a single molecule.

### Step 3: Substitute the average velocity into the momentum formula
Substituting \(v_{\text{average}}\) into the momentum formula gives:
\[ p_{\text{average}} = m \cdot \sqrt{\frac{8RT}{\pi m}} \]
This simplifies to:
\[ p_{\text{average}} = \sqrt{8RTm/\pi} \]

### Step 4: Gather the values from the problem
From the problem, we have:
- Temperature \(T = 150 \pi K\)
- Mass of a helium molecule \(m = \frac{166}{3} \times 10^{-27} \text{ kg}\)
- Universal gas constant \(R = \frac{25}{3} \text{ J mol}^{-1} \text{ K}^{-1}\)

### Step 5: Substitute the values into the momentum formula
Now, substituting the values into the formula:
\[ p_{\text{average}} = \sqrt{\frac{8 \cdot \left(\frac{25}{3}\right) \cdot (150 \pi) \cdot \left(\frac{166}{3} \times 10^{-27}\right)}{\pi}} \]
This simplifies to:
\[ p_{\text{average}} = \sqrt{\frac{8 \cdot \frac{25}{3} \cdot 150 \cdot 166 \times 10^{-27}}{3}} \]

### Step 6: Calculate the numerical value
Calculating the expression inside the square root:
1. Calculate \(8 \cdot \frac{25}{3} \cdot 150 \cdot 166\):
\[ = 8 \cdot \frac{25 \cdot 150 \cdot 166}{3} \]
= \(8 \cdot \frac{625000}{3} = \frac{5000000}{3}\)

2. Now, substituting this back into the square root:
\[ p_{\text{average}} = \sqrt{\frac{5000000 \times 10^{-27}}{3}} = \sqrt{\frac{5000000}{3}} \times 10^{-13.5} \]

3. Finally, calculate the square root and simplify:
\[ p_{\text{average}} \approx 3 \times 10^{-23} \text{ kg m/s} \]

### Final Answer
The average magnitude of linear momentum of helium molecules is approximately:
\[ p_{\text{average}} \approx 3 \times 10^{-23} \text{ kg m/s} \]

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