The Earth's atmosphere consists primarily of oxygen `(21%)` and nitrogen `(78%)`. The rms speed of oxygen molecules `(O_(2))` in the atmosphere at a certain location is 535 m/s. The rms speed of the nitrogen molecules at this location will be [Given `"1 amu "=1.66xx10^(-26)" kg, molecular mass of "O_(2)=32," molecular mass of "N_(2)=28]`

Options

A. 572 m/s
B. 437 m/s
C. 835 m/s
D. 715 m/s

Answer

To find the root mean square (rms) speed of nitrogen molecules in the atmosphere, we can use the formula for rms speed:

\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]

where:
- \( R \) is the universal gas constant, \( R = 8.314 \, \text{J/(mol K)} \)
- \( T \) is the temperature in Kelvin
- \( M \) is the molar mass in kg/mol

### Step 1: Calculate the Temperature (T)

We know the rms speed of oxygen \( (O_2) \) is given as 535 m/s. We can use this information to find the temperature \( T \).

Using the rms speed formula for oxygen:

\[ v_{\text{rms, O2}} = \sqrt{\frac{3RT}{M_{O2}}} \]

Substituting the known values:

\[ 535 = \sqrt{\frac{3 \times 8.314 \times T}{32 \times 10^{-3}}} \]

### Step 2: Square Both Sides

To eliminate the square root, we square both sides:

\[ 535^2 = \frac{3 \times 8.314 \times T}{32 \times 10^{-3}} \]

Calculating \( 535^2 \):

\[ 287225 = \frac{3 \times 8.314 \times T}{0.032} \]

### Step 3: Rearranging the Equation

Now, rearranging the equation to solve for \( T \):

\[ T = \frac{287225 \times 0.032}{3 \times 8.314} \]

### Step 4: Calculate the Temperature (T)

Calculating \( T \):

\[ T = \frac{9171.2}{24.942} \approx 367.21 \, \text{K} \]

### Step 5: Calculate the rms Speed of Nitrogen (N2)

Now we will use the temperature \( T \) to find the rms speed of nitrogen \( (N_2) \):

\[ v_{\text{rms, N2}} = \sqrt{\frac{3RT}{M_{N2}}} \]

Substituting the values:

\[ v_{\text{rms, N2}} = \sqrt{\frac{3 \times 8.314 \times 367.21}{28 \times 10^{-3}}} \]

### Step 6: Calculate the rms Speed of Nitrogen (N2)

Calculating \( v_{\text{rms, N2}} \):

\[ v_{\text{rms, N2}} = \sqrt{\frac{3 \times 8.314 \times 367.21}{0.028}} \]

Calculating the numerator:

\[ 3 \times 8.314 \times 367.21 \approx 9171.2 \]

Now substituting back:

\[ v_{\text{rms, N2}} = \sqrt{\frac{9171.2}{0.028}} \approx \sqrt{327,000} \approx 571.93 \, \text{m/s} \]

### Step 7: Round the Answer

Rounding to a reasonable figure, we get:

\[ v_{\text{rms, N2}} \approx 572 \, \text{m/s} \]

### Final Answer

The rms speed of nitrogen molecules at this location is approximately **572 m/s**.

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