`0.040g` of He is kept in a closed container initially at `100.0^(@)C.`The container is now heated. Neglecting the expansion of the container, Calculate the temperature at which the internal energy is increased by `12J`.

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Identify the properties of helium
Helium is a monoatomic gas. For monoatomic gases, the specific heat at constant volume \( C_v \) is given by:
\[ C_v = \frac{3}{2} R \]
where \( R \) is the universal gas constant, approximately \( 8.31 \, \text{J/(mol K)} \).

### Step 2: Convert the mass of helium to moles
The mass of helium is given as \( 0.040 \, \text{g} \). The molar mass of helium is \( 4 \, \text{g/mol} \). To find the number of moles (\( n \)):
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.040 \, \text{g}}{4 \, \text{g/mol}} = 0.01 \, \text{mol} \]

### Step 3: Convert the initial temperature to Kelvin
The initial temperature is given as \( 100.0^\circ C \). To convert this to Kelvin:
\[ T_1 = 100 + 273 = 373 \, \text{K} \]

### Step 4: Use the internal energy formula
The change in internal energy (\( \Delta U \)) for an ideal gas is given by:
\[ \Delta U = n C_v \Delta T \]
where \( \Delta T = T_2 - T_1 \).

### Step 5: Substitute known values into the internal energy equation
We know \( \Delta U = 12 \, \text{J} \), \( n = 0.01 \, \text{mol} \), and \( C_v = \frac{3}{2} R \). First, calculate \( C_v \):
\[ C_v = \frac{3}{2} \times 8.31 \, \text{J/(mol K)} = 12.465 \, \text{J/(mol K)} \]

Now substitute into the internal energy equation:
\[ 12 = 0.01 \times 12.465 \times (T_2 - 373) \]

### Step 6: Solve for \( T_2 \)
Rearranging the equation:
\[ 12 = 0.12465 (T_2 - 373) \]
\[ T_2 - 373 = \frac{12}{0.12465} \]
Calculating the right side:
\[ T_2 - 373 = 96.32 \]
Thus,
\[ T_2 = 96.32 + 373 = 469.32 \, \text{K} \]

### Step 7: Convert \( T_2 \) back to Celsius
To convert the final temperature from Kelvin to Celsius:
\[ T_2 = 469.32 - 273 = 196.32^\circ C \]

### Final Answer
The temperature at which the internal energy is increased by \( 12 \, \text{J} \) is approximately:
\[ T_2 \approx 196.32^\circ C \]