For one mole of an ideal gas, increasing the temperature from `10^(@)C` to `20^(@)C`

Options

A. increases the average kinetic energy by two times
B. increases the rms velocity of `sqrt2` times
C. increases the rms velocity by two times
D. increases both the average kinetic energy and rms velocity, but not significantly

Answer

To solve the problem of determining the effect of increasing the temperature of one mole of an ideal gas from \(10^\circ C\) to \(20^\circ C\), we will follow these steps:

### Step 1: Convert Celsius to Kelvin
To analyze the behavior of the gas, we first need to convert the temperatures from Celsius to Kelvin. The conversion formula is:
\[ T(K) = T(°C) + 273 \]
For \(10^\circ C\):
\[ T_1 = 10 + 273 = 283 \, K \]
For \(20^\circ C\):
\[ T_2 = 20 + 273 = 293 \, K \]

### Step 2: Calculate Average Kinetic Energy
The average kinetic energy (\(E\)) of an ideal gas is given by the formula:
\[ E = \frac{3}{2} k T \]
where \(k\) is the Boltzmann constant and \(T\) is the temperature in Kelvin.

For \(E_1\) at \(T_1\):
\[ E_1 = \frac{3}{2} k (283) \]
For \(E_2\) at \(T_2\):
\[ E_2 = \frac{3}{2} k (293) \]

### Step 3: Find the Ratio of Average Kinetic Energies
To find the ratio of the average kinetic energies \(E_2\) to \(E_1\):
\[ \frac{E_2}{E_1} = \frac{\frac{3}{2} k (293)}{\frac{3}{2} k (283)} = \frac{293}{283} \]
Calculating this gives:
\[ \frac{E_2}{E_1} \approx 1.0354 \]
This indicates that the average kinetic energy increases by approximately \(3.54\%\).

### Step 4: Calculate Root Mean Square Velocity
The root mean square velocity (\(v_{rms}\)) is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass.

For \(v_{rms1}\) at \(T_1\):
\[ v_{rms1} = \sqrt{\frac{3R(283)}{M}} \]
For \(v_{rms2}\) at \(T_2\):
\[ v_{rms2} = \sqrt{\frac{3R(293)}{M}} \]

### Step 5: Find the Ratio of Root Mean Square Velocities
To find the ratio of \(v_{rms2}\) to \(v_{rms1}\):
\[ \frac{v_{rms2}}{v_{rms1}} = \frac{\sqrt{3R(293)/M}}{\sqrt{3R(283)/M}} = \sqrt{\frac{293}{283}} \]
Calculating this gives:
\[ \frac{v_{rms2}}{v_{rms1}} \approx \sqrt{1.0354} \approx 1.0176 \]
This indicates that the root mean square velocity increases by approximately \(1.76\%\).

### Conclusion
From the calculations:
- The average kinetic energy increases by approximately \(3.54\%\).
- The root mean square velocity increases by approximately \(1.76\%\).

### Final Answer
Both the average kinetic energy and the root mean square velocity increase, but not by a factor of 2. The increases are approximately \(1.0354\) times for kinetic energy and \(1.0176\) times for root mean square velocity.