`(1//2)` mole of helium is contained in a container at STP how much heat energy is needed to double the pressure of the gas, keeping the volume constant? Heat capacity of gas is `3 J g^(-1)K^(-1)`.

Options

A. 1436 J
B. 736 J
C. 1638 J
D. 5698 J

Explanation

To solve the problem of how much heat energy is needed to double the pressure of helium gas while keeping the volume constant, we can follow these steps:

### Step 1: Understand the relationship between pressure and temperature
At constant volume, the pressure of a gas is directly proportional to its absolute temperature (Kelvin). This relationship can be expressed as:
\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \]

### Step 2: Set up the initial conditions
Given that we want to double the pressure:
\[ P_2 = 2P_1 \]
Thus, we can write:
\[ \frac{2P_1}{P_1} = \frac{T_2}{T_1} \]
This simplifies to:
\[ 2 = \frac{T_2}{T_1} \]
From this, we can find:
\[ T_2 = 2T_1 \]

### Step 3: Calculate the change in temperature
The change in temperature (\( \Delta T \)) is given by:
\[ \Delta T = T_2 - T_1 = 2T_1 - T_1 = T_1 \]

### Step 4: Convert the initial temperature to Kelvin
At standard temperature and pressure (STP), the temperature is:
\[ T_1 = 273 \, \text{K} \]
Thus, the change in temperature becomes:
\[ \Delta T = T_1 = 273 \, \text{K} \]

### Step 5: Calculate the heat energy required
The heat energy (\( \Delta Q \)) required to change the temperature of the gas can be calculated using the formula:
\[ \Delta Q = n C_V \Delta T \]
Where:
- \( n \) is the number of moles of the gas,
- \( C_V \) is the heat capacity at constant volume,
- \( \Delta T \) is the change in temperature.

Given:
- \( n = \frac{1}{2} \, \text{mole} \)
- \( C_V = 3 \, \text{J/g/K} \)

We need to convert moles to grams. The molar mass of helium (He) is approximately \( 4 \, \text{g/mol} \):
\[ \text{Mass of helium} = n \times \text{Molar mass} = \frac{1}{2} \times 4 \, \text{g} = 2 \, \text{g} \]

Now substituting the values into the heat equation:
\[ \Delta Q = \left(\frac{1}{2}\right) \times 3 \, \text{J/g/K} \times 273 \, \text{K} \]
\[ \Delta Q = \frac{1}{2} \times 3 \times 273 \]
\[ \Delta Q = \frac{819}{2} \]
\[ \Delta Q = 409.5 \, \text{J} \]

### Final Calculation
Thus, the total heat energy required to double the pressure of the gas at constant volume is:
\[ \Delta Q = 409.5 \, \text{J} \]