The rms velocity of a gas at T °C is double the value at 27 °C. The temperature T of the gas in °C is ( assume that the pressure remains constant)

Options

A. 927
B. 820
C. 1000
D. 195

Answer

To find the temperature \( T \) of the gas in °C, given that the root mean square (rms) velocity of the gas at \( T \) °C is double the value at 27 °C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the RMS Velocity Formula**:
The rms velocity \( v_{\text{rms}} \) 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, and \( M \) is the molar mass of the gas.

2. **Convert Temperatures to Kelvin**:
- The temperature at 27 °C in Kelvin is:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]
- The temperature \( T \) in Celsius can be converted to Kelvin as:
\[ T_2 = T + 273 \, \text{K} \]

3. **Set Up the Equation**:
According to the problem, the rms velocity at \( T \) °C is double that at 27 °C:
\[ v_{\text{rms}}(T) = 2 \cdot v_{\text{rms}}(27) \]
Substituting the rms velocity formula:
\[ \sqrt{\frac{3R(T + 273)}{M}} = 2 \cdot \sqrt{\frac{3R(300)}{M}} \]

4. **Square Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ \frac{3R(T + 273)}{M} = 4 \cdot \frac{3R(300)}{M} \]
The \( 3R/M \) terms cancel out:
\[ T + 273 = 4 \cdot 300 \]

5. **Calculate**:
Simplifying the equation:
\[ T + 273 = 1200 \]
Therefore:
\[ T = 1200 - 273 \]
\[ T = 927 \, \text{°C} \]

### Final Answer:
The temperature \( T \) of the gas is \( 927 \, \text{°C} \).