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 and Intuition

As a bubble rises, the surrounding water pressure decreases (less depth of water above it) while temperature stays constant, so by Boyle's law $PV=\text{const}$, meaning $P_1 r_1^3 = P_2 r_2^3$ (since $V\propto r^3$). Pressure at any depth is atmospheric pressure plus the pressure due to the water column above that point.

Step-by-Step Solution

1. Pressure at the bottom (depth 7.28 m): $P_1 = 10 + 7.28 = 17.28$ m of water (in head units).
2. Pressure at the top surface: $P_2 = 10$ m of water.
3. Boyle's law: $P_1 r_1^3 = P_2 r_2^3 \Rightarrow \dfrac{r_2^3}{r_1^3} = \dfrac{P_1}{P_2} = \dfrac{17.28}{10} = 1.728$. …

Concept and Intuition

A soap bubble has two liquid surfaces, so its excess (gauge) pressure is $P=\dfrac{4T}{r}$. In vacuum, this excess pressure IS the absolute pressure of the enclosed gas. When two bubbles merge, the enclosed gas (moles) is conserved and the process is isothermal, so by the ideal gas law $PV = nRT$ stays additive: the $PV$ of the merged bubble equals the sum of the $PV$'s of the two original bubbles.

Step-by-Step Solution

1. For a bubble of radius $r$: $P = \dfrac{4T}{r}$, $V=\dfrac43\pi r^3$, so $PV = \dfrac{4T}{r}\times\dfrac43\pi r^3 = \dfrac{16\pi T}{3}r^2$ — proportional to $r^2$.
2. Conservation of total gas amount (isothermal, same temperature, no gas lost) means $P_{\text{new}}V_{\text{new}} = P_1V_1+P_2V_2$, i.e. $R^2 = r_1^2+r_2^2$.
3. With $r_1=r_2=2\,$cm: $R^2 = 2^2+2^2 = 8 \Rightarrow R = \sqrt8 = 2\sqrt2\,\text{cm}$.

Common Mistakes …

Concept and Intuition

Because the vessel's volume and the amount of gas inside are both fixed, this is a constant-volume process, governed by Gay-Lussac's law: pressure is directly proportional to absolute temperature. As the vessel is heated, the pressure climbs steadily until it reaches the vessel's structural limit of 100 atm, at which point it explodes.

Step-by-Step Solution

1. Initial state: $P_1 = 20$ atm at $T_1 = 27°C = 300$ K.
2. Constant-volume law: $\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$.
3. The vessel explodes when $P_2 = 100$ atm (its withstand limit). …

Concept and Intuition

For an ideal gas at constant volume, $PV=nRT \Rightarrow P=\left(\dfrac{nR}{V}\right)T$, i.e. $P\propto T$ (an exactly proportional/linear relation, not just approximately, since $V$ is fixed). Any percentage increase in one produces the identical percentage increase in the other.

Step-by-Step Solution

1. From $PV=nRT$ with $V$, $n$ fixed: $P=kT$ for constant $k=nR/V$.
2. If $P$ increases by 2%: $P'=1.02P=k(1.02T)$, so $T'=1.02T$. …

Concept and Intuition

For a fixed amount of ideal gas, $\dfrac{PV}{T}$ is constant (from $PV=nRT$ with $n,R$ fixed), so any change in volume/temperature/pressure can be related directly through the combined gas law, without needing to compute the number of moles explicitly.

Step-by-Step Solution

1. Convert temperatures to Kelvin: $T_1=27+273=300$K, $T_2=T_1+30=330$K.
2. Combined gas law: $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$.
3. Substitute known values: $\dfrac{12\times10}{300}=\dfrac{P_2\times6}{330}$.
4. LHS $=\dfrac{120}{300}=0.4$. …

Concept and Intuition

At constant temperature, for a fixed amount of ideal gas, pressure and volume are inversely proportional: $PV=\text{constant}$. Lowering the pressure must increase the volume proportionally.

Step-by-Step Solution

1. Initial state: $P_1=760\ \text{mm Hg}$, $V_1=100\ \text{cc}$.
2. Final state: $P_2=400\ \text{mm Hg}$, $V_2=?$ (same temperature).
3. Boyle's law: $P_1V_1 = P_2V_2 \implies 760\times100 = 400\times V_2$. …

Concept and Intuition

$\gamma=C_p/C_v$ only matters for an adiabatic process, where $PV^\gamma=$ constant. The question explicitly says the compression is isothermal, so temperature (hence $PV=nRT=$ constant) governs the process — Boyle's law applies, and $\gamma$ is a distractor.

Step-by-Step Solution

1. Isothermal process: $P_1V_1=P_2V_2$.
2. Given $P_1=1.0\ atm$, $V_2=\dfrac{V_1}{2}$.
3. $P_2=\dfrac{P_1V_1}{V_2}=\dfrac{P_1V_1}{V_1/2}=2P_1=2.0\ atm$.

Common Mistakes …

Concept and Intuition

As the bubble rises, both the pressure and temperature it experiences change: pressure decreases (less water column above it) and temperature increases (surface is warmer). Both effects act to increase the volume: lower pressure lets the gas expand, and higher temperature also expands it. The combined gas law $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$ (valid for a fixed amount of gas obeying the ideal-gas law) lets us combine both effects in one equation.

Step-by-Step Solution

1. Pressure at the bottom = atmospheric + hydrostatic pressure of the water column: $P_1 = P_0 + \rho g h$, with $P_0=1.01\times10^5$ Pa, $\rho=1000\ kg\,m^{-3}$, $g=9.8\ m\,s^{-2}$, $h=40\ m$.
2. $\rho g h = 1000\times9.8\times40 = 3.92\times10^5$ Pa, so $P_1 = 1.01\times10^5+3.92\times10^5 = 4.93\times10^5$ Pa.
3. At the surface, pressure is just atmospheric: $P_2 = 1.01\times10^5$ Pa.
4. Convert temperatures to kelvin: $T_1 = 285\ K$, $T_2 = 308\ K$. …

Concept and Intuition

For an ideal gas at constant pressure, volume is directly proportional to absolute (kelvin) temperature — this is Charles's law, a special case of the ideal gas law $PV=nRT$ with $P$ held fixed. So doubling the absolute temperature doubles the volume, and so on. The key step most students miss is converting the given Celsius temperatures into kelvin before taking the ratio.

Step-by-Step Solution

1. Convert temperatures to kelvin: $T_1 = 27+273=300\,K$; $T_2=297+273=570\,K$.
2. Apply Charles's law: $\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} \Rightarrow V_2 = V_1\times\dfrac{T_2}{T_1}$. …

Concept and Intuition

Charles's law states that at constant pressure, volume is directly proportional to absolute temperature: $V \propto T$. This is a direct consequence of the ideal gas law $PV=nRT$ with $P$ and $n$ fixed. The crucial step is always converting Celsius to Kelvin before using the ratio, since the proportionality only holds for absolute temperature.

Step-by-Step Solution

1. Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300\ K$, $T_2 = -3 + 273 = 270\ K$.
2. Apply Charles's law: $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$. …

Concept and Intuition

When a fixed amount of gas changes pressure, volume, and temperature simultaneously, the combined gas law $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ (equivalent to the ideal gas law with $n$, $R$ constant) lets us relate the two states directly without needing to know the actual number of moles.

Step-by-Step Solution

1. Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300\ \text{K}$, $T_2 = -3 + 273 = 270\ \text{K}$.
2. Given: $P_1 = 4\ \text{atm}$, $V_1 = 1500\ \text{m}^3$; $P_2 = 2\ \text{atm}$.
3. Combined gas law: $V_2 = V_1 \times \dfrac{P_1}{P_2} \times \dfrac{T_2}{T_1}$. …

Concept and Intuition

The combined gas law $\dfrac{PV}{T}=\text{const}$ (for a fixed amount of gas) says temperature scales directly with the product $PV$. If both $P$ and $V$ double, $PV$ becomes four times as large, so $T$ must become four times as large too.

Step-by-Step Solution

1. Initial: $P_1=2\,Pa$, $T_1=27^\circ C=300\,K$ (some volume $V_1$).
2. Final: $P_2=2P_1=4\,Pa$, $V_2=2V_1$. …