The mass 15 gram of Nitrogen is enclosed in vessel at 300K. What heat must be supplies to it to double the 'rms' velocity of its molecules

Options

A. 10J
B. 10KJ
C. `10^(3)J`
D. `10^(2)J`

Explanation

To solve the problem of how much heat must be supplied to double the root mean square (rms) velocity of nitrogen molecules, we can follow these steps:

### Step 1: Understand the relationship between rms velocity and temperature
The rms velocity (\(v_{rms}\)) 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 2: Set up the equation for doubling the rms velocity
If we want to double the rms velocity, we can set up the equation:
\[ 2v_{rms1} = v_{rms2} \]
Substituting the rms velocity formula, we have:
\[ 2\sqrt{\frac{3RT_1}{M}} = \sqrt{\frac{3RT_2}{M}} \]
Squaring both sides gives:
\[ 4\frac{3RT_1}{M} = \frac{3RT_2}{M} \]
Canceling \(3R/M\) from both sides leads to:
\[ 4T_1 = T_2 \]

### Step 3: Calculate the final temperature
Given that the initial temperature \(T_1\) is 300 K:
\[ T_2 = 4 \times 300 = 1200 \, \text{K} \]

### Step 4: Calculate the change in internal energy
The change in internal energy (\(\Delta U\)) for an ideal gas can be expressed as:
\[ \Delta U = nC_v\Delta T \]
where:
- \(n\) is the number of moles,
- \(C_v\) is the molar specific heat at constant volume,
- \(\Delta T = T_2 - T_1\).

### Step 5: Calculate the number of moles
The mass of nitrogen is given as 15 grams. The molar mass of nitrogen (\(N_2\)) is approximately 28 g/mol. Thus, the number of moles \(n\) is:
\[ n = \frac{m}{M} = \frac{15 \, \text{g}}{28 \, \text{g/mol}} \approx 0.536 \, \text{mol} \]

### Step 6: Determine \(C_v\)
For diatomic gases like nitrogen, the molar specific heat at constant volume \(C_v\) is given by:
\[ C_v = \frac{R}{\gamma - 1} \]
where \(\gamma = \frac{C_p}{C_v}\) and for diatomic gases, \(\gamma \approx \frac{7}{5}\). Thus:
\[ C_v = \frac{R}{\frac{7}{5} - 1} = \frac{5R}{2} \]

### Step 7: Calculate \(\Delta T\)
The change in temperature \(\Delta T\) is:
\[ \Delta T = T_2 - T_1 = 1200 - 300 = 900 \, \text{K} \]

### Step 8: Calculate the heat supplied (\(\Delta Q\))
Since the work done (\(\Delta W\)) is zero in a closed vessel, the heat supplied is equal to the change in internal energy:
\[ \Delta Q = \Delta U = nC_v\Delta T \]
Substituting the values:
\[ \Delta Q = 0.536 \times \frac{5 \times 8.31}{2} \times 900 \]
Calculating \(C_v\):
\[ C_v \approx \frac{5 \times 8.31}{2} = 20.775 \, \text{J/mol K} \]
Now substituting:
\[ \Delta Q = 0.536 \times 20.775 \times 900 \]
Calculating:
\[ \Delta Q \approx 1004 \, \text{J} \approx 10 \, \text{kJ} \]

### Final Answer
The heat that must be supplied to double the rms velocity of the nitrogen molecules is approximately **10 kJ**.

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