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 & Intuition

When an air bubble rises in a lake, the temperature is constant, so we apply Boyle’s law: $P_1 V_1 = P_2 V_2$. The pressure at any depth is the sum of atmospheric pressure (given as equivalent to 10 m of water) plus the pressure due to the water column above that depth. As the bubble rises, the surrounding pressure decreases, so the volume increases. The question asks for the percentage increase between two specific depths, not from the bottom.

Step-by-step solution

1. Express pressures in metres of water column Atmospheric pressure = pressure of 10 m of water. At a depth $h$ metres, the total pressure is:

2. Find pressures at the two depths At depth 20 m:

At depth 15 m:

3. Apply Boyle’s law Since temperature is constant: …

Concept and intuition

We have a fixed-volume vessel. Initially it contains a certain mass of gas at pressure $P$ and temperature $127^\circ\mathrm{C}$. After some gas leaks out, the pressure drops to $\frac{2P}{3}$ and the temperature changes to an unknown $t$. The mass that leaked out is given, so we know the mass remaining. Since the volume and the gas are the same (same molar mass), the ideal gas law $PV = nRT$ tells us that the number of moles $n$ is proportional to $P/T$. Mass is proportional to moles, so the mass of gas inside the vessel is proportional to $P/T$. This lets us set up a ratio without needing the volume or the gas constant.

Step-by-step solution

1. Convert temperatures to Kelvin

   Initial temperature: $127^\circ\mathrm{C} = 127 + 273 = 400\,\mathrm{K}$.

   Final temperature: $t^\circ\mathrm{C} = (t + 273)\,\mathrm{K}$. We'll call this $T$.

2. Determine the mass remaining after leakage

   Initial mass = $21\,\mathrm{g}$. Mass leaked out = $5\,\mathrm{g}$.

   So mass remaining in the vessel = $21 - 5 = 16\,\mathrm{g}$.

3. Apply the ideal gas law for constant volume

   For a fixed volume $V$ and a fixed gas (same molar mass $M$), the number of moles $n = \frac{m}{M}$ satisfies

Rearranging:

Since $V$, $M$, and $R$ are constants, mass $m$ is proportional to $\frac{P}{T}$.

4. Set up the ratio Let initial state: $P_1 = P$, $T_1 = 400\,\mathrm{K}$, $m_1 = 21\,\mathrm{g}$. Final state (after leakage): $P_2 = \frac{2P}{3}$, $T_2 = T$, $m_2 = 16\,\mathrm{g}$. From proportionality: $$$$

The key here is that an air bubble in water is not at constant pressure — the deeper you go, the more the water above presses down. As the bubble rises, the surrounding pressure drops, so the bubble expands. On top of that, the temperature at the surface is a bit higher than at the bottom, which also makes the gas expand. Both effects work together.

We treat the air inside the bubble as an ideal gas. The number of moles stays fixed (no air leaks in or out), so we can use the combined gas law:

We know the initial volume $V_1 = 0.7$ cc. We need to find the pressure at the bottom and at the surface, convert temperatures to Kelvin, then solve for $V_2$, and finally find the increase $V_2 - V_1$.

1. Pressure at the bottom of the lake

   The total pressure at depth $h$ is the sum of atmospheric pressure and the pressure due to the water column:

Given: $P_{\text{atm}} = 1 \times 10^5$ Pa, $g = 10$ m/s², $h = 10$ m. The density of water $\rho = 1000$ kg/m³.

So:

2. Pressure at the surface

   At the surface, the bubble is only under atmospheric pressure:

3. Convert temperatures to Kelvin

   Bottom temperature: $T_1 = 7^\circ\text{C} = 7 + 273 = 280$ K

   Surface temperature: $T_2 = 11^\circ\text{C} = 11 + 273 = 284$ K

4. Apply the combined gas law

Rearranging for $V_2$:

Substitute:

Simplify $\frac{284}{280} = \frac{71}{70}$:

$V_2 = 0.7 \times 2 \times \frac{71}{70} = 0.7 \times \frac{142}{70}$ …

The key idea here is that the coefficient of volume expansion (often denoted $\gamma$ or $\beta$) measures how much a substance’s volume changes per degree change in temperature, relative to its original volume. For an ideal gas at constant pressure, this coefficient is not a material constant — it depends directly on the temperature itself. That’s because the ideal gas law $PV = nRT$ tells us that at fixed pressure, volume is proportional to absolute temperature: $V \propto T$. So a 1% increase in absolute temperature gives a 1% increase in volume, and the proportionality constant is simply $1/T$.

Let’s work through it step by step.

1. Recall the definition The coefficient of volume expansion at constant pressure is defined as

It tells us the fractional change in volume per unit temperature change.

2. Apply the ideal gas law For $n$ moles of an ideal gas at constant pressure $P$:

Here $T$ must be in kelvins (absolute temperature). The derivative is straightforward:

3. Substitute into the definition

So for an ideal gas at constant pressure, $\gamma = 1/T$. This is a clean, memorable result.

4. Convert the given temperature to kelvins The problem gives $27^\circ\text{C}$.

Hence

Why this approach works

An adiabatic process is one in which no heat enters or leaves the system. When you compress a gas adiabatically, all the work you do on it goes into increasing its internal energy, which for an ideal gas means raising its temperature. The relationship between temperature and volume during such a process comes from combining the ideal gas law with the first law of thermodynamics and the condition $Q = 0$.

This tells us that as volume decreases, temperature must increase to keep the product constant.

Solution

1. Convert initial temperature to Kelvin

   The initial temperature is $T_1 = 27\,^\circ\mathrm{C} = 27 + 273 = 300\,\mathrm{K}$.

2. Identify the volume ratio

   The gas is compressed to $\frac{8}{27}$ of its initial volume, so:

3. Calculate $\gamma - 1$

   With $\gamma = \frac{5}{3}$:

4. Apply the adiabatic relation

   From $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$, we can write:

Substituting the values:

5. Evaluate the power …

The key idea is that any temperature scale that is linear (like Celsius, Fahrenheit, or this new X scale) can be converted using a simple proportion based on two fixed reference points. Here, the melting point of ice and the boiling point of water are the same physical temperatures on both scales, so they define the endpoints of a straight-line relationship.

We know:

• On Celsius: ice point = $0^\circ\mathrm{C}$, steam point = $100^\circ\mathrm{C}$.
• On X scale: ice point = $20^\circ\mathrm{X}$, steam point = $110^\circ\mathrm{X}$.

The difference between these fixed points on Celsius is $100^\circ$, and on X scale it is $110 - 20 = 90^\circ$. So every $1^\circ\mathrm{C}$ corresponds to $90/100 = 0.9^\circ\mathrm{X}$.

But we also need to account for the offset: $0^\circ\mathrm{C}$ corresponds to $20^\circ\mathrm{X}$, not $0^\circ\mathrm{X}$.

1. Set up the linear relationship. If $C$ is the temperature in Celsius and $X$ is the temperature on the new scale, then

where $m$ is the slope and $b$ is the intercept.

2. Use the two fixed points to find $m$ and $b$.
   • At $C = 0$, $X = 20$:

• At $C = 100$, $X = 110$:

3. Write the conversion formula.

4. Plug in $C = 40$. $$$$

Concept & Intuition

When a bubble rises, the water pressure decreases, so the bubble expands. Since temperature is constant, we use Boyle’s law: $P_1 V_1 = P_2 V_2$. The pressure at the bottom is the sum of atmospheric pressure (expressed as height $H$ of water column) plus the pressure due to the water depth $h$. At the surface, only atmospheric pressure acts. The volume becomes 5 times larger, so the pressure must become 5 times smaller — but careful: it’s the product that stays constant, so if volume goes up by 5, pressure goes down by 5. That gives a direct equation for $h$.

Step-by-step solution

1. Define pressures Let atmospheric pressure be $P_{\text{atm}} = \rho g H$ (where $\rho$ is water density, $g$ gravity, and $H$ the height of water column that produces the same pressure). At the bottom of the lake, depth $h$ adds pressure $\rho g h$. So total pressure at bottom:

At the surface, pressure is just atmospheric:

2. Apply Boyle’s law Since temperature is constant:

Given $V_{\text{surface}} = 5 V_{\text{bottom}}$, substitute:

$\rho g (H + h) \cdot V_{\text{bottom}} = \rho g H \cdot 5 V_{\text{bottom}}$ …

Concept & Intuition

A soap bubble has two liquid-air interfaces — an inner surface and an outer surface — each contributing to the surface energy. Work done in expanding the bubble goes entirely into increasing this surface energy (since the process is isothermal and we ignore other losses). The key formula is:

Because the bubble has two surfaces, the total area is twice the spherical surface area.

Step-by-step solution

1. Identify the radii

   Initial radius $r_1 = 1~\text{cm} = 0.01~\text{m}$

   Final radius $r_2 = 4~\text{cm} = 0.04~\text{m}$

2. Surface area of one spherical surface

   Area of a sphere: $4\pi r^2$

   Initial one-surface area: $4\pi (0.01)^2 = 4\pi \times 10^{-4}~\text{m}^2$

   Final one-surface area: $4\pi (0.04)^2 = 4\pi \times 16 \times 10^{-4} = 64\pi \times 10^{-4}~\text{m}^2$

3. Total surface area of the bubble (two surfaces)

   Initial total area: $A_1 = 2 \times 4\pi r_1^2 = 8\pi (0.01)^2 = 8\pi \times 10^{-4}~\text{m}^2$

   Final total area: $A_2 = 2 \times 4\pi r_2^2 = 8\pi (0.04)^2 = 8\pi \times 16 \times 10^{-4} = 128\pi \times 10^{-4}~\text{m}^2$

4. Change in total surface area

5. Work done …

The key concept here is the first law of thermodynamics combined with the definition of an adiabatic process.

• Adiabatic means no heat exchange with the surroundings: $Q = 0$.
• The first law says $\Delta U = Q - W$, where $W$ is the work done by the gas.
• So for adiabatic expansion: $\Delta U = -W$. Since the gas expands, it does positive work ($W > 0$), so $\Delta U < 0$.
• A decrease in internal energy for an ideal gas means a decrease in temperature ($T$).
• For expansion, volume increases; from the ideal gas law $PV = nRT$, if $T$ drops and $V$ rises, pressure $P$ must also drop.

Let’s walk through it step by step.

1. Identify the process: Adiabatic expansion — volume increases, no heat transfer ($Q = 0$).

2. Apply the first law:

Since the gas expands, $W > 0$, so $\Delta U < 0$. Internal energy decreases.

3. Relate internal energy to temperature: For an ideal gas, internal energy depends only on temperature:

With $\Delta U < 0$, we get $\Delta T < 0$ — temperature decreases.

4. Use the ideal gas law to see pressure change:

During expansion, $V$ increases. We just found $T$ decreases. For the product $P V$ to equal $n R T$ (which is smaller), $P$ must also decrease. (If $P$ increased, $P V$ would increase, contradicting the drop in $T$.) …

The key idea is that for most metals, resistance changes approximately linearly with temperature over moderate ranges. The formula is

where $R_0$ is the resistance at $0\,^\circ\mathrm{C}$, $\alpha$ is the temperature coefficient of resistance, and $T$ is the temperature in $^\circ\mathrm{C}$. Once we determine $\alpha$ from the known resistances at $0\,^\circ\mathrm{C}$ and $40\,^\circ\mathrm{C}$, we can plug in the new resistance to find the unknown temperature.

1. Find the temperature coefficient $\alpha$. At $0\,^\circ\mathrm{C}$, $R_0 = 4\,\Omega$. At $40\,^\circ\mathrm{C}$, $R_{40} = 4.06\,\Omega$. Using the linear relation:

Substitute:

Divide both sides by 4:

So

2. Set up the equation for the bath temperature. Let the bath temperature be $T$. The resistance in the bath is $4.15\,\Omega$, so:

Divide by 4:

Thus

$\alpha T = 0.0375$ …