Question

3.) a gas at 3.00l and 2.00atm expands until the pressure is 1.00atm what is the final volume?

Explanation:

Step1: Identify the Gas Law

This is a Boyle's Law problem (since temperature is constant, \( P_1V_1 = P_2V_2 \)).
Given \( P_1 = 2.00 \, \text{atm} \), \( V_1 = 3.00 \, \text{L} \), \( P_2 = 1.00 \, \text{atm} \), find \( V_2 \).

Step2: Apply Boyle's Law Formula

Boyle's Law: \( P_1V_1 = P_2V_2 \)
Rearrange to solve for \( V_2 \): \( V_2 = \frac{P_1V_1}{P_2} \)

Step3: Substitute Values

\( P_1 = 2.00 \, \text{atm} \), \( V_1 = 3.00 \, \text{L} \), \( P_2 = 1.00 \, \text{atm} \)
\( V_2 = \frac{(2.00 \, \text{atm})(3.00 \, \text{L})}{1.00 \, \text{atm}} \)

Step4: Calculate

\( V_2 = \frac{6.00 \, \text{L·atm}}{1.00 \, \text{atm}} = 6.00 \, \text{L} \)

Answer:

The final volume is \( \boldsymbol{6.00 \, \text{L}} \)