The temperature of an ideal gas is increased from `27 ^@ C` to `927^(@)C`. The rms speed of its molecules becomes.

Options

A. twice
B. half
C. four times
D. one - fourth

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The initial temperature \( T_1 \) is given as \( 27^\circ C \) and the final temperature \( T_2 \) is \( 927^\circ C \). To convert these temperatures to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Calculating:
\[ T_1 = 27 + 273 = 300 \, K \]
\[ T_2 = 927 + 273 = 1200 \, K \]

### Step 2: Understand the relationship between RMS speed and temperature
The root mean square (RMS) speed \( v_{rms} \) of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant and \( M \) 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 \( T \):
\[ v_{rms} \propto \sqrt{T} \]

### Step 3: Set up the ratio of RMS speeds
Since the RMS speed is proportional to the square root of the temperature, we can write the ratio of the initial and final RMS speeds as:
\[ \frac{v_{rms1}}{v_{rms2}} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]

### Step 4: Substitute the temperatures into the ratio
Substituting the values of \( T_1 \) and \( T_2 \):
\[ \frac{v_{rms1}}{v_{rms2}} = \frac{\sqrt{300}}{\sqrt{1200}} = \frac{\sqrt{300}}{\sqrt{1200}} = \frac{\sqrt{300}}{\sqrt{4 \times 300}} = \frac{\sqrt{300}}{2\sqrt{300}} = \frac{1}{2} \]

### Step 5: Solve for the final RMS speed
From the ratio, we find:
\[ \frac{v_{rms1}}{v_{rms2}} = \frac{1}{2} \]
This implies:
\[ v_{rms2} = 2 \times v_{rms1} \]
Thus, the RMS speed of the gas molecules doubles when the temperature is increased from \( 27^\circ C \) to \( 927^\circ C \).

### Final Answer
The final RMS speed of the gas molecules becomes twice the initial speed.

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