Question

A bubble of air is underwater at temperature $15^{\circ}C$ and the pressure $1.5 \,bar$. If the bubble rises to the surface where the temperature is $25^{\circ}C$ and the pressure is $1.0\, bar$, what will happen to the volume of the bubble?

• A. Volume will become greater by a factor of 2.5
• B. Volume will become greater by a factor of 1.6
• C. Volume will become greater by a factor of 1.1
• D. Volume will become smaller by a factor of 0.70

Answer 1

The Correct Option is B

To solve this problem, we can use the Combined Gas Law, which relates the pressure, volume, and temperature of a gas. The Combined Gas Law is given by:

\frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2}

Where:

• P_1, V_1, T_1 are the initial pressure, volume, and temperature, respectively.
• P_2, V_2, T_2 are the final pressure, volume, and temperature, respectively.

Let's substitute the given values:

• P_1 = 1.5 \, \text{bar}, T_1 = 15^{\circ}\text{C} = 15 + 273.15 = 288.15 \, \text{K}
• P_2 = 1.0 \, \text{bar}, T_2 = 25^{\circ}\text{C} = 25 + 273.15 = 298.15 \, \text{K}

Assuming the initial volume of the bubble is V_1, we aim to find the final volume V_2:

\frac{1.5 \cdot V_1}{288.15} = \frac{1.0 \cdot V_2}{298.15}

Rearrange to solve for V_2:

V_2 = V_1 \cdot \left(\frac{1.5}{1.0}\right) \cdot \left(\frac{298.15}{288.15}\right)

Now calculate the volume change factor:

V_2 = V_1 \cdot \left(1.5\right) \cdot \left(\frac{298.15}{288.15}\right) \approx V_1 \cdot 1.6

Thus, the volume of the bubble will be greater by a factor of approximately 1.6.

Therefore, the correct option is: Volume will become greater by a factor of 1.6.