Question

A bubble of an ideal gas rises from the bottom of a lake to the surface. At the bottom, the pressure is 3 Atm. and the temperature is 7 °C. At the surface, the pressure is 1 atm. and the temperature is 27 °C. If the initial volume of the bubble was $V_0$ what is its volume $V_f$ at the surface?

• A. $3 V_0$
• B. $3.21 V_0$
• C. $0.9 V_0$
• D. $5.4 V_0$

Hint:

Always convert Celsius to Kelvin in gas law problems by adding 273.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
As the bubble rises, the pressure decreases and the temperature changes. We use the Combined Gas Law to relate the initial state (bottom) to the final state (surface). It is critical to convert temperatures to Kelvin.
Step 2: Key Formula or Approach:
1. Combined Gas Law: \( \frac{P_1 V_1}{T_1} = \frac{P_f V_f}{T_f} \).
2. Temperature conversion: \( T(K) = T(^\circ C) + 273 \).
Step 3: Detailed Explanation:
Initial state (bottom): \( P_1 = 3 \text{ atm} \), \( V_1 = V_0 \), \( T_1 = 7 + 273 = 280 \text{ K} \).
Final state (surface): \( P_f = 1 \text{ atm} \), \( T_f = 27 + 273 = 300 \text{ K} \).
Rearranging for \( V_f \): \[ V_f = \frac{P_1 V_1 T_f}{T_1 P_f} \] \[ V_f = \frac{3 \times V_0 \times 300}{280 \times 1} \] \[ V_f = \frac{900}{280} V_0 \approx 3.214 V_0 \]
Step 4: Final Answer:
The volume at the surface is 3.21 \( V_0 \).