A compressor pumps `70 L` of air into a `6 L` tank with the tempertaure remaining unchanged. If all the air is originally at `1 atm`. What is the fianl absolute pressure of the air in the tank ?

Explanation

To solve the problem, we can use the ideal gas law principle, particularly the relationship between pressure and volume when temperature is constant. This is derived from the equation \( PV = nRT \), which can be simplified to \( P_1V_1 = P_2V_2 \) when the amount of gas (n) and the gas constant (R) are unchanged, and temperature remains constant.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial volume of air, \( V_1 = 70 \, L \)
- Initial pressure of air, \( P_1 = 1 \, atm \)
- Final volume of the tank, \( V_2 = 6 \, L \)

2. **Apply the Ideal Gas Law**:
- Since the temperature is constant, we can use the relationship \( P_1V_1 = P_2V_2 \).

3. **Rearranging the Equation**:
- We need to find the final pressure \( P_2 \). Rearranging the equation gives:
\[ P_2 = \frac{P_1V_1}{V_2} \]

4. **Substituting the Values**:
- Now, substitute the known values into the equation:
\[ P_2 = \frac{(1 \, atm) \times (70 \, L)}{6 \, L} \]

5. **Calculating \( P_2 \)**:
- Performing the calculation:
\[ P_2 = \frac{70}{6} \, atm \approx 11.67 \, atm \]

6. **Conclusion**:
- The final absolute pressure of the air in the tank is approximately \( 11.67 \, atm \).