Kinetic energy of 0.30 moles of He gas in a container of maximum capacity of 4 litres at 5 atmosphere, must be (R = 0.0821 atm litre `"mole"^(-1) K^(-1)` ) ________`xx 10^(1)` atm.lit

Answer

To solve the problem of finding the kinetic energy of 0.30 moles of helium gas in a container of maximum capacity of 4 liters at 5 atmospheres, we can follow these steps:

### Step 1: Use the Ideal Gas Law to find the Temperature (T)
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = Pressure (in atm)
- \( V \) = Volume (in liters)
- \( n \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Temperature (in Kelvin)

Given:
- \( P = 5 \, \text{atm} \)
- \( V = 4 \, \text{liters} \)
- \( n = 0.30 \, \text{moles} \)
- \( R = 0.0821 \, \text{atm} \cdot \text{litre} \cdot \text{mole}^{-1} \cdot \text{K}^{-1} \)

Rearranging the ideal gas law to solve for \( T \):
\[ T = \frac{PV}{nR} \]

Substituting the values:
\[ T = \frac{(5 \, \text{atm}) \times (4 \, \text{liters})}{(0.30 \, \text{moles}) \times (0.0821 \, \text{atm} \cdot \text{litre} \cdot \text{mole}^{-1} \cdot \text{K}^{-1})} \]

### Step 2: Calculate the Temperature (T)
Calculating the numerator:
\[ 5 \times 4 = 20 \]

Calculating the denominator:
\[ 0.30 \times 0.0821 = 0.02463 \]

Now substituting these values into the equation for \( T \):
\[ T = \frac{20}{0.02463} \approx 812.4 \, \text{K} \]

### Step 3: Calculate the Kinetic Energy (KE)
The kinetic energy of an ideal gas is given by the formula:
\[ KE = \frac{3}{2} nRT \]

Substituting the values:
\[ KE = \frac{3}{2} \times 0.30 \times 0.0821 \times 812.4 \]

### Step 4: Calculate the Kinetic Energy (KE)
Calculating \( nRT \):
\[ nRT = 0.30 \times 0.0821 \times 812.4 \approx 20.1 \]

Now substituting this back into the kinetic energy formula:
\[ KE = \frac{3}{2} \times 20.1 \approx 30.15 \]

### Step 5: Express the Kinetic Energy in the Required Format
The final answer can be expressed in the form \( xx \times 10^1 \):
\[ KE \approx 30.15 \approx 3.015 \times 10^1 \]

Thus, rounding it off, we can say:
\[ KE \approx 3 \times 10^1 \, \text{atm.lit} \]

### Final Answer
The kinetic energy of 0.30 moles of He gas in the given conditions is approximately:
\[ \boxed{3} \]