At constant pressure, calculate the root mean square velocity of a gas molecule at temperature `27^@ C` if its rms speed at `0^@ C`. Is 4 km/s.

Explanation

To solve the problem of finding the root mean square (rms) velocity of a gas molecule at a temperature of 27°C, given that its rms speed at 0°C is 4 km/s, we can follow these steps:

### Step 1: Understand the relationship between rms speed and temperature
The root mean square speed (vrms) of a gas is given by the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
Where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

From this formula, we can see that the rms speed is directly proportional to the square root of the temperature. Therefore, we can write the relationship as:
\[ \frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{T_1}{T_2}} \]

### Step 2: Convert temperatures from Celsius to Kelvin
- For \( T_1 = 27°C \):
\[ T_1 = 27 + 273 = 300 \, K \]
- For \( T_2 = 0°C \):
\[ T_2 = 0 + 273 = 273 \, K \]

### Step 3: Substitute the known values into the equation
We know:
- \( v_{\text{rms2}} = 4 \, \text{km/s} \) (at 0°C)
- \( T_1 = 300 \, K \)
- \( T_2 = 273 \, K \)

Now we can substitute these values into the equation:
\[ \frac{v_{\text{rms1}}}{4} = \sqrt{\frac{300}{273}} \]

### Step 4: Solve for \( v_{\text{rms1}} \)
To find \( v_{\text{rms1}} \), we rearrange the equation:
\[ v_{\text{rms1}} = 4 \cdot \sqrt{\frac{300}{273}} \]

### Step 5: Calculate the square root and final value
Now we calculate the square root:
\[ \sqrt{\frac{300}{273}} \approx \sqrt{1.0985} \approx 1.048 \]
Then, substituting back:
\[ v_{\text{rms1}} \approx 4 \cdot 1.048 \approx 4.192 \, \text{km/s} \]

### Final Answer
Thus, the root mean square velocity of the gas molecule at 27°C is approximately:
\[ \boxed{4.19 \, \text{km/s}} \]