A diver releases a bubble of air of volume 8.00\text{ cm}^3 at a depth where the water pre… — JAMB UTME

A diver releases a bubble of air of volume $8.00\text{ cm}^3$ at a depth where the water pressure is $3.50 \times 10^5\text{ Pa}$ and the temperature is $7^\circ\text{C}$ . What is the volume of the air bubble, in $\text{cm}^3$ , just as it reaches the surface where the pressure is $1.00 \times 10^5\text{ Pa}$ and the temperature is $27^\circ\text{C}$ ?

Answer: 30 cm³

Answer

The final volume of the air bubble at the surface is

30.0\text{ cm}^3

Using the combined gas law

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

with absolute temperatures

T_1 = 280\text{ K}

and

T_2 = 300\text{ K}

yields a final volume of

30.0\text{ cm}^3

Step-by-Step Solution

Convert temperatures from Celsius to Kelvin

T_1 = 7^\circ\text{C} + 273 = 280\text{ K}

and

T_2 = 27^\circ\text{C} + 273 = 300\text{ K}

Gas laws require absolute thermodynamic temperature in Kelvin.

Apply the combined gas law equation

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

The amount of gas in the bubble remains constant while pressure, volume, and temperature all change simultaneously.

Rearrange for the unknown volume

V_2

and substitute the values

V_2 = \frac{3.50 \times 10^5 \times 8.00 \times 300}{1.00 \times 10^5 \times 280} = 30.0\text{ cm}^3

Calculates the expanded volume of the air bubble at surface conditions.

Key Concept

Combined Gas Law

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