Air is cooled from `25^@C` to `0^@C`. Calculate the decrease in rms speed of the molecules.

Options

A. `4 %`
B. `12 %`
C. `29 %`
D. `50 %`

Answer

To calculate the decrease in the root mean square (rms) speed of air molecules when the temperature is cooled from \(25^\circ C\) to \(0^\circ C\), we can follow these steps:

### Step 1: Convert Temperatures to Kelvin
The temperatures need to be converted from Celsius to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
- For \(T_1 = 25^\circ C\):
\[ T_1 = 25 + 273 = 298 \, K \]
- For \(T_2 = 0^\circ C\):
\[ T_2 = 0 + 273 = 273 \, K \]

### Step 2: Write the Formula for rms Speed
The rms speed \(v\) of gas molecules is given by the formula:
\[ v = \sqrt{\frac{3RT}{M}} \]
where:
- \(R\) is the universal gas constant,
- \(T\) is the temperature in Kelvin,
- \(M\) is the molar mass of the gas.

### Step 3: Calculate Initial and Final rms Speeds
- For the initial speed \(v_1\) at \(T_1\):
\[ v_1 = \sqrt{\frac{3R \cdot 298}{M}} \]
- For the final speed \(v_2\) at \(T_2\):
\[ v_2 = \sqrt{\frac{3R \cdot 273}{M}} \]

### Step 4: Calculate the Percentage Decrease in rms Speed
The percentage decrease in rms speed can be calculated using the formula:
\[ \text{Percentage Decrease} = \frac{v_1 - v_2}{v_1} \times 100 \]
Substituting the expressions for \(v_1\) and \(v_2\):
\[ \text{Percentage Decrease} = \frac{\sqrt{\frac{3R \cdot 298}{M}} - \sqrt{\frac{3R \cdot 273}{M}}}{\sqrt{\frac{3R \cdot 298}{M}}} \times 100 \]
The common terms \(\sqrt{\frac{3R}{M}}\) cancel out:
\[ = \frac{\sqrt{298} - \sqrt{273}}{\sqrt{298}} \times 100 \]

### Step 5: Calculate the Numerical Values
Now, we compute the numerical values:
- Calculate \(\sqrt{298} \approx 17.26\)
- Calculate \(\sqrt{273} \approx 16.52\)

Substituting these values:
\[ \text{Percentage Decrease} = \frac{17.26 - 16.52}{17.26} \times 100 \approx \frac{0.74}{17.26} \times 100 \approx 4.29\% \]

### Step 6: Final Calculation
Thus, the percentage decrease in rms speed is approximately \(4.29\%\), which can be rounded to \(4\%\).

### Final Answer
The decrease in rms speed of the molecules is approximately **4%**.

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