Volume of the air bubble, V_{1} = 1.0 cm^{3}

= 1.0 x 10^{-6 }m^{3}

Air bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T_{1} = 12°C

= 285 K

Temperature at the surface of the lake, T_{2 }= 35°C

= 308 K

The pressure on the surface of the lake:

P_{2} = 1 atm = 1 x 1.013 x 10^{5} Pa

The pressure at the depth of 40 m:

P_{1}= 1 atm + dρg

Where,

ρ is the density of water = 10^{3} kg / m^{3}

g is the acceleration due to gravity = 9.8 m/s^{2}

Hence,

P_{1} = 1.013 x 10^{5} + 40 x 10^{3} x 9.8

We get,

= 493300 Pa

We have

\( \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)

Where, V_{2} is the volume of the air bubble when it reaches the surface

\(V2 = \dfrac{P_1V_1T_2}{T_1P_2}\)

= \( \dfrac{493300 \times 1 \times 10^{-6} \times 308 }{(285 \times 1.013 \times 10^5)}\)

We get,

= 5.263 x 10^{-6} m^{3} or 5.263 cm^{3}

Hence, when the air bubble reaches the surface, its volume becomes 5.263 cm^{3}

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance |
Atomic Mass (u) |
Kg m^{-3}) |

Carbon (diamond) | 12.01 | 2.22 |

Gold | 197.00 | 19.32 |

Nitrogen (liquid) | 14.01 | 1.00 |

Lithium | 6.94 | 0.53 |

Fluorine (liquid) | 19.00 | 1.14 |

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n_{2}= n_{1} exp [ -mg \(
\dfrac{(h_2–h_1)}{k_BT}\)T] where n_{2}, n_{1 }refer to number density at heights h_{2 }and h_{1 }respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:n_{2} = n_{1} exp [ -mg N_{A} (ρ – ρ′ ) \(
\dfrac{(h_2–h_1)}{ρ RT}\)] where ρ is the density of the suspended particle, and ρ′ ,that of surrounding medium. [N_{A} is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]

From a certain apparatus, the diffusion rate of hydrogen has an average value of \( 28.7 cm^3 s^{-1}\). The diffusion of another gas under the same conditions is measured to have an average rate of \( 7.2 cm^3 s^{-1}\). Identify the gas. [Hint: Use Graham’s law of diffusion: \( \dfrac{R_1}{R_2}\) = \( (\dfrac{M_2}{M_1})^ \dfrac{1}{2}\), where \( R_1 , R_2\) are diffusion rates of gases 1 and 2, and \( M_1\) and \( M_2\) their respective molecular masses. The law is a simple consequence of kinetic theory.]

A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 170°C . Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of \( N_2\) = 28.0 u).