Questions: 1 point 5.30 atm of air in a 2.85 L container is being decompressed to fit into a custom diving tank. Assuming the temperature of the gas remains constant, what would be the pressure of the gas if it was transferred into a 4.50 L container? Round your answer to the nearest hundredth. Type your answer... Previous Next

1 point 5.30 atm of air in a 2.85 L container is being decompressed to fit into a custom diving tank. Assuming the temperature of the gas remains constant, what would be the pressure of the gas if it was transferred into a 4.50 L container? Round your answer to the nearest hundredth. Type your answer... Previous Next
Transcript text: 1 point 5.30 atm of air in a 2.85 L container is being decompressed to fit into a custom diving tank. Assuming the temperature of the gas remains constant, what would be the pressure of the gas if it was transferred into a 4.50 L container? Round your answer to the nearest hundredth. Type your answer... Previous Next
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Solution

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Solution Steps

Step 1: Identify the Given Values

We are given:

  • Initial pressure, \( P_1 = 5.30 \) atm
  • Initial volume, \( V_1 = 2.85 \) L
  • Final volume, \( V_2 = 4.50 \) L
Step 2: Apply Boyle's Law

Boyle's Law states that for a given mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, this is expressed as: \[ P_1 V_1 = P_2 V_2 \]

Step 3: Solve for the Final Pressure

Rearrange the equation to solve for \( P_2 \): \[ P_2 = \frac{P_1 V_1}{V_2} \]

Substitute the given values into the equation: \[ P_2 = \frac{5.30 \, \text{atm} \times 2.85 \, \text{L}}{4.50 \, \text{L}} \]

Step 4: Calculate the Final Pressure

Perform the calculation: \[ P_2 = \frac{15.105 \, \text{atm} \cdot \text{L}}{4.50 \, \text{L}} \approx 3.3567 \, \text{atm} \]

Step 5: Round to the Nearest Hundredth

Round the result to the nearest hundredth: \[ P_2 \approx 3.36 \, \text{atm} \]

Final Answer

\(\boxed{P_2 = 3.36 \, \text{atm}}\)

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