A cylinder of fixed capacity `44.8` litres constains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by `15^(@)C` ? Given `R=8.31 j mol e^(-1) K^(-1)` . (For monoatomic gas, `C_(v)=3 R//2`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the volume of helium gas to moles
Given that the volume of the cylinder is \(44.8\) liters and at standard temperature and pressure (STP), \(1\) mole of gas occupies \(22.4\) liters. We can calculate the number of moles (\(N\)) of helium gas in the cylinder using the formula:

\[ N = \frac{\text{Volume}}{\text{Volume per mole}} = \frac{44.8 \, \text{liters}}{22.4 \, \text{liters/mole}} = 2 \, \text{moles} \]

### Step 2: Determine the specific heat capacity at constant volume (\(C_V\))
For a monoatomic gas like helium, the specific heat capacity at constant volume is given by:

\[ C_V = \frac{3}{2} R \]

Where \(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\). Thus,

\[ C_V = \frac{3}{2} \times 8.31 = 12.465 \, \text{J mol}^{-1} \text{K}^{-1} \]

### Step 3: Calculate the heat required to raise the temperature
The amount of heat (\(Q\)) required to raise the temperature of the gas can be calculated using the formula:

\[ Q = N C_V \Delta T \]

Where:
- \(N\) is the number of moles (which we found to be \(2\)),
- \(C_V\) is the specific heat capacity (which we calculated as \(12.465 \, \text{J mol}^{-1} \text{K}^{-1}\)),
- \(\Delta T\) is the change in temperature, which is \(15 \, \text{°C}\) (or \(15 \, \text{K}\) since the change in Celsius is equivalent to the change in Kelvin).

Substituting the values we have:

\[ Q = 2 \, \text{moles} \times 12.465 \, \text{J mol}^{-1} \text{K}^{-1} \times 15 \, \text{K} \]

Calculating this gives:

\[ Q = 2 \times 12.465 \times 15 = 374 \, \text{J} \]

### Final Answer
The amount of heat needed to raise the temperature of the gas in the cylinder by \(15 \, \text{°C}\) is \(374 \, \text{J}\).

---