The total kinetic energy of 8 litres of helium molecules at 5 atmosophere pressure will be (1 atmosphere `=1.013xx10^(5)` pascal)

Options

A. 607J
B. 6078J
C. 607erg
D. 6078erg

Explanation

To find the total kinetic energy of 8 liters of helium molecules at 5 atmospheres of pressure, we can use the formula for the kinetic energy of an ideal gas:

\[ KE = \frac{3}{2} nRT \]

where:
- \( KE \) is the total kinetic energy,
- \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant (\( R = 8.314 \, \text{J/(mol K)} \)),
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Convert the volume from liters to cubic meters
Given that \( 1 \, \text{liter} = 0.001 \, \text{m}^3 \):

\[ V = 8 \, \text{liters} = 8 \times 0.001 \, \text{m}^3 = 0.008 \, \text{m}^3 \]

### Step 2: Convert pressure from atmospheres to pascals
Given that \( 1 \, \text{atmosphere} = 1.013 \times 10^5 \, \text{Pa} \):

\[ P = 5 \, \text{atmospheres} = 5 \times 1.013 \times 10^5 \, \text{Pa} = 5.065 \times 10^5 \, \text{Pa} \]

### Step 3: Use the ideal gas law to find the number of moles \( n \)
The ideal gas law is given by:

\[ PV = nRT \]

Rearranging for \( n \):

\[ n = \frac{PV}{RT} \]

We need to know the temperature \( T \). Assuming standard room temperature, \( T \approx 298 \, \text{K} \):

Substituting the values:

\[ n = \frac{(5.065 \times 10^5 \, \text{Pa})(0.008 \, \text{m}^3)}{(8.314 \, \text{J/(mol K)})(298 \, \text{K})} \]

Calculating \( n \):

\[ n = \frac{4052.8}{2477.572} \approx 1.63 \, \text{moles} \]

### Step 4: Calculate the total kinetic energy
Now we can substitute \( n \), \( R \), and \( T \) back into the kinetic energy formula:

\[ KE = \frac{3}{2} nRT = \frac{3}{2} (1.63 \, \text{moles})(8.314 \, \text{J/(mol K)})(298 \, \text{K}) \]

Calculating \( KE \):

\[ KE = \frac{3}{2} \times 1.63 \times 8.314 \times 298 \approx \frac{3}{2} \times 4060.1 \approx 2454.1 \, \text{J} \]

### Final Answer
The total kinetic energy of 8 liters of helium molecules at 5 atmospheres of pressure is approximately:

\[ KE \approx 2454.1 \, \text{J} \]