Q.An air bubble of volume $1.0\ \text{cm}^3$ rises from the bottom of a lake $40\ \text{m}$ deep at a temperature of $12\ ^\circ\text{C}$. To what volume does it grow when it reaches the surface, which is at a temperature of $35\ ^\circ\text{C}$?

Concept: Combined Gas Law

The bubble experiences changing pressure (depth) and temperature as it rises. Assuming the amount of gas is constant and it behaves ideally, we use:

Step 1: Identify conditions at the bottom and surface.

At 40 m depth: $P_1 = P_{\text{atm}} + \rho g h = 1.013 \times 10^5 + (1000)(10)(40) = 5.013 \times 10^5\ \text{Pa}$, $V_1 = 1.0\ \text{cm}^3$, $T_1 = 12 + 273 = 285\ \text{K}$.

At surface: $P_2 = 1.013 \times 10^5\ \text{Pa}$, $T_2 = 35 + 273 = 308\ \text{K}$.

Step 2: Solve for $V_2$. …

Why the combined gas law?

An air bubble rising through water experiences two simultaneous changes: the pressure decreases as the weight of water above it diminishes, and the temperature increases as it moves toward the warmer surface. Since the amount of gas (number of moles) remains constant, we need a relationship that connects pressure, volume, and temperature for a fixed quantity of gas.

The combined gas law does exactly this:

This emerges directly from the ideal gas equation $PV = nRT$. When $n$ and $R$ are constant, the ratio $\frac{PV}{T}$ must remain constant between any two states.

Step-by-step solution

1. Identify the initial state (at the bottom)

The bubble starts at a depth of $40\ \text{m}$ below the surface. The pressure at this depth is the sum of atmospheric pressure and the pressure due to the water column above:

where $P_0 \approx 10^5\ \text{Pa}$ is atmospheric pressure, $\rho = 1000\ \text{kg/m}^3$ is the density of water, and $g = 10\ \text{m/s}^2$.

The initial volume is $V_1 = 1.0\ \text{cm}^3$ and the temperature is $T_1 = 12 + 273 = 285\ \text{K}$.

2. Identify the final state (at the surface)

At the surface, the bubble experiences only atmospheric pressure:

The temperature at the surface is $T_2 = 35 + 273 = 308\ \text{K}$.

We need to find $V_2$.

3. Apply the combined gas law

Substituting into $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$:

$\frac{5P_0 \times 1.0}{285} = \frac{P_0 \times V_2}{308}$ …

Shortcut: separate the two effects and multiply. Since $V_2 = V_1 \cdot \dfrac{P_1}{P_2} \cdot \dfrac{T_2}{T_1}$, you don't need to solve the combined law as one block — compute the pressure ratio and the temperature ratio separately, then multiply them into $V_1$. Here $P_1/P_2 = 5P_0/P_0 = 5$ (pure Boyle's-law expansion) and $T_2/T_1 = 308/285 \approx 1.08$ (pure Charles's-law expansion), so $V_2 \approx 1.0 \times 5 \times 1.08 = 5.4\ \text{cm}^3$. This factoring doubles as a built-in sanity check: the …