The temperature of a gas is - `68^(@)C`. To what temperature should it be heated, so that the r.m.s. velocity of the molecules be doubled ?

Options

A. `357^(@)C`
B. `457^(@)C`
C. `547^(@)`
D. `820^(@)`

Explanation

To solve the problem, we need to understand the relationship between the root mean square (r.m.s.) velocity of gas molecules and temperature. The r.m.s. velocity \( v_{rms} \) is given by the formula:

\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]

where:
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the mass of a gas molecule.

Since we want to double the r.m.s. velocity, we can set up the following relationship:

\[ v'_{rms} = 2v_{rms} \]

Substituting the formula for r.m.s. velocity, we have:

\[ \sqrt{\frac{3kT'}{m}} = 2\sqrt{\frac{3kT}{m}} \]

Squaring both sides gives:

\[ \frac{3kT'}{m} = 4 \cdot \frac{3kT}{m} \]

The \( 3k/m \) terms cancel out, leading to:

\[ T' = 4T \]

Now, we need to find the initial temperature \( T \) in Kelvin. The initial temperature is given as \( -68^\circ C \). To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Calculating the initial temperature:

\[ T = -68 + 273.15 = 205.15 \, K \]

Now, we can find the new temperature \( T' \):

\[ T' = 4T = 4 \times 205.15 = 820.6 \, K \]

Finally, we convert \( T' \) back to Celsius:

\[ T' (°C) = T' (K) - 273.15 = 820.6 - 273.15 = 547.45 \, °C \]

Thus, the temperature to which the gas should be heated is approximately:

\[ \boxed{547.45 \, °C} \]