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 of finding the change in the root mean square (rms) speed of the molecules of an ideal gas when the temperature is increased from \(27^\circ C\) to \(927^\circ C\), we can follow these steps:

### Step 1: Convert the temperatures to Kelvin
The first step is to convert the given temperatures from Celsius to Kelvin, as the ideal gas law and related equations use absolute temperature.

\[ T_1 = 27^\circ C = 27 + 273 = 300 \, K \]
\[ T_2 = 927^\circ C = 927 + 273 = 1200 \, K \]

### Step 2: Use the formula for rms speed
The rms speed (\(V_{rms}\)) of gas molecules is given by the formula:

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

Where:
- \(R\) is the universal gas constant,
- \(T\) is the absolute temperature in Kelvin,
- \(M\) is the molar mass of the gas.

### Step 3: Understand the relationship between rms speed and temperature
From the formula, we can see that the rms speed is directly proportional to the square root of the temperature:

\[ V_{rms} \propto \sqrt{T} \]

### Step 4: Calculate the ratio of rms speeds at the two temperatures
Let \(V_1\) be the rms speed at \(T_1\) and \(V_2\) be the rms speed at \(T_2\). The ratio of the rms speeds can be expressed as:

\[ \frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}} \]

### Step 5: Substitute the values of \(T_1\) and \(T_2\)
Now, substitute the values of \(T_1\) and \(T_2\):

\[ \frac{V_2}{V_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2 \]

### Step 6: Conclude the relationship between \(V_2\) and \(V_1\)
From the ratio calculated, we find:

\[ V_2 = 2V_1 \]

This means that the rms speed of the gas molecules at \(927^\circ C\) is twice that at \(27^\circ C\).

### Final Answer
Thus, the rms speed of the molecules becomes twice as much when the temperature is increased from \(27^\circ C\) to \(927^\circ C\).

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