An air bubble with an initial volume of 4.0\text{ cm}^3 is released by a scuba diver at a … — JAMB UTME

An air bubble with an initial volume of $4.0\text{ cm}^3$ is released by a scuba diver at a depth where the total pressure is $2.50 \times 10^5\text{ Pa}$ and the water temperature is $7^\circ\text{C}$ . Calculate the volume of the bubble, in $\text{cm}^3$ , just as it reaches the surface where the pressure is $1.00 \times 10^5\text{ Pa}$ and the water temperature is $77^\circ\text{C}$ .

Answer: 12.5 cm³

Answer

12.5 cm³

Converting temperatures to the Kelvin scale (

T_1 = 280\text{ K}

and

T_2 = 350\text{ K}

) and applying the combined gas law formula

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

gives a final volume of

12.5\text{ cm}^3

Step-by-Step Solution

Convert the initial and final temperatures from degrees Celsius to absolute temperatures in Kelvin.

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

and

T_2 = 77^\circ\text{C} + 273 = 350\text{ K}

Gas law calculations require thermodynamic temperature measured on the Kelvin scale.

Set up the combined gas law relationship for a fixed mass of gas.

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

The mass of air inside the bubble remains constant while pressure, volume, and temperature change simultaneously.

Substitute the known values into the equation and solve for the final volume

V_2

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{(2.50 \times 10^5\text{ Pa}) \times (4.0\text{ cm}^3) \times (350\text{ K})}{(1.00 \times 10^5\text{ Pa}) \times (280\text{ K})} = 12.5\text{ cm}^3

Algebraic substitution yields the correct final volume of the expanded bubble.

Key Concept

Combined Gas Law

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