A gas bulb of 1 litre capacity contains `2.0 xx 10^(21)` molecules of nitrogen exerting a pressure of `7.57 xx 10^3 Nm^(-2)` . Calculate the root mean square speed and the temperature of gas molecules. If the ratio of most probable speed to the root mean square speed is 0.82, calculate the most probable speed for the molecules at this temperature.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the volume from liters to cubic meters.
Given that the volume of the gas bulb is 1 liter, we convert it to cubic meters:
\[ \text{Volume} = 1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3 \]

### Step 2: Calculate the number of moles of nitrogen gas.
We know the number of molecules of nitrogen gas is \(2.0 \times 10^{21}\). To find the number of moles, we use Avogadro's number (\(6.022 \times 10^{23} \, \text{molecules/mol}\)):
\[ n = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{2.0 \times 10^{21}}{6.022 \times 10^{23}} \approx 0.00332 \, \text{mol} \]

### Step 3: Use the ideal gas equation to find the temperature.
The ideal gas equation is given by:
\[ PV = nRT \]
Rearranging for temperature \(T\):
\[ T = \frac{PV}{nR} \]
Substituting the values:
- Pressure \(P = 7.57 \times 10^{3} \, \text{N/m}^2\)
- Volume \(V = 1 \times 10^{-3} \, \text{m}^3\)
- Number of moles \(n = 0.00332 \, \text{mol}\)
- Gas constant \(R = 8.314 \, \text{J/(mol K)}\)

Now substituting:
\[ T = \frac{(7.57 \times 10^{3}) \times (1 \times 10^{-3})}{0.00332 \times 8.314} \approx 274.25 \, \text{K} \]

### Step 4: Calculate the root mean square (RMS) speed of the gas molecules.
The formula for the root mean square speed \(v_{rms}\) is:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Where \(M\) is the molar mass in kg. For nitrogen (\(N_2\)), the molar mass is \(28 \, \text{g/mol} = 0.028 \, \text{kg/mol}\).

Substituting the values:
\[ v_{rms} = \sqrt{\frac{3 \times 8.314 \times 274.25}{0.028}} \approx 403.53 \, \text{m/s} \]

### Step 5: Calculate the most probable speed.
Given that the ratio of the most probable speed \(v_{mp}\) to the root mean square speed \(v_{rms}\) is 0.82:
\[ \frac{v_{mp}}{v_{rms}} = 0.82 \]
Thus,
\[ v_{mp} = 0.82 \times v_{rms} = 0.82 \times 403.53 \approx 330.89 \, \text{m/s} \]

### Final Answers:
- Root Mean Square Speed: \(v_{rms} \approx 403.53 \, \text{m/s}\)
- Temperature: \(T \approx 274.25 \, \text{K}\)
- Most Probable Speed: \(v_{mp} \approx 330.89 \, \text{m/s}\)

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