A mass M = 15 g of nitrogen is enclosed in a vessel at temperature T = 300 K. What amount of heat has to be transferred to the gas to increase the root-mean-square velocity of molecules 2 times ?

Options

A. `10 kJ`
B. `6 kJ`
C. `0.9 kJ`
D. `14 kJ`

Explanation

To solve the problem, we need to determine the amount of heat that must be transferred to a mass of nitrogen gas in order to double the root-mean-square (RMS) velocity of its molecules. Here’s a step-by-step solution:

### Step 1: Understanding the relationship between RMS velocity and temperature
The root-mean-square velocity (v_rms) of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas in kg/mol.

From this formula, we can see that \( v_{rms} \) is proportional to the square root of the temperature \( T \).

### Step 2: Setting up the equation for doubling the RMS velocity
If we want to double the RMS velocity, we set up the equation:
\[ 2v_{rms} = \sqrt{\frac{3R(4T)}{M}} \]
This implies that to double \( v_{rms} \), we need to increase the temperature to \( 4T \).

### Step 3: Calculate the initial and final temperatures
Given that the initial temperature \( T_0 = 300 \, K \), the final temperature \( T_f \) required to double the RMS velocity is:
\[ T_f = 4T_0 = 4 \times 300 = 1200 \, K \]

### Step 4: Calculate the change in temperature
The change in temperature \( \Delta T \) is:
\[ \Delta T = T_f - T_0 = 1200 - 300 = 900 \, K \]

### Step 5: Calculate the number of moles of nitrogen
The molar mass of nitrogen (N₂) is approximately 28 g/mol. The mass \( M \) given is 15 g. Therefore, the number of moles \( n \) is:
\[ n = \frac{M}{\text{Molar mass}} = \frac{15 \, g}{28 \, g/mol} = \frac{15}{28} \, mol \approx 0.536 \, mol \]

### Step 6: Calculate the change in internal energy
For a diatomic gas, the degrees of freedom \( f = 5 \). The change in internal energy \( \Delta U \) for an ideal gas can be calculated using:
\[ \Delta U = \frac{f}{2} n R \Delta T \]
Substituting the values:
\[ \Delta U = \frac{5}{2} \times 0.536 \, mol \times 8.314 \, J/(mol \cdot K) \times 900 \, K \]

### Step 7: Calculate the total heat transferred
Calculating \( \Delta U \):
\[ \Delta U = \frac{5}{2} \times 0.536 \times 8.314 \times 900 \]
\[ \Delta U \approx \frac{5}{2} \times 0.536 \times 7482.6 \approx 10,000 \, J \]

### Step 8: Final answer
The amount of heat that has to be transferred to the gas to increase the RMS velocity of the molecules 2 times is approximately:
\[ Q \approx 10,000 \, J \text{ or } 10 \, kJ \]