Answer

Diameter of an oxygen molecule, $d=3 \mathring{A}$

Radius, $r=\frac{d}{2}=\frac{3}{2}=1.5 \mathring{A}=1.5 \times 10^{-8} cm$

Actual volume occupied by 1 mole of oxygen gas at $STP=22400 cm^{3}$

Molecular volume of oxygen gas,

$ V=\frac{4}{3} \pi r^{3} \cdot N $

Where, $N$ is Avogadro’s number $=6.023 \times 10^{23}$ molecules $/ mole$

$\therefore V=\frac{4}{3} \times 3.14 \times(1.5 \times 10^{-8})^{3} \times 6.023 \times 10^{23}=8.51 cm^{3}$

Ratio of the molecular volume to the actual volume of oxygen $=\frac{8.51}{22400}$

$=3.8 \times 10^{-4}$

Answer

The ideal gas equation relating pressure $(P)$, volume $(V)$, and absolute temperature $(T)$ is given as: $P V=n R T$

Where,

$R$ is the universal gas constant $=8.314 J mol^{-1} K^{-1}$

$n=$ Number of moles $=1$

$T=$ Standard temperature $=273 K$

$P=$ Standard pressure $=1 atm=1.013 \times 10^{5} Nm^{-2}$

$\therefore V=\frac{n R T}{P}$

$=\frac{1 \times 8.314 \times 273}{1.013 \times 10^{5}}$

$=0.0224 m^{3}$

$=22.4$ litres

Hence, the molar volume of a gas at STP is 22.4 litres.

Answer

(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio $\frac{P V}{T}$ is equal. $\mu R$ ( $\mu$ is the number of moles and $R$ is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.

(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature $T_1$ is closer to the dotted plot than the curve of the gas at temperature $T_2$. A real gas approaches the behaviour of an ideal gas when its temperature increases.

Therefore, $T_1>T_2$ is true for the given plot.

(c) The value of the ratio $P V / T$, where the two curves meet, is $\mu R$. This is because the ideal gas equation is given as:

$P V=\mu R T$

$\frac{P V}{T}=\mu R$

Where,

$P$ is the pressure

$T$ is the temperature

$V$ is the volume

$\mu$ is the number of moles

$R$ is the universal constant

Molecular mass of oxygen $=32.0 g$

Mass of oxygen $=1 \times 10^{-3} kg=1 g$

$R=8.314 J mole^{-1} K^{-1}$

$\therefore \frac{P V}{T}=\frac{1}{32} \times 8.314$ $=0.26 J K^{-1}$

Therefore, the value of the ratio $P V / T$, where the curves meet on the $y$-axis, is

$0.26 J K^{-1}$.

(d) If we obtain similar plots for $1.00 \times 10^{-3} kg$ of hydrogen, then we will not get the same value of $P V / T$ at the point where the curves meet the $y$-axis. This is because the molecular mass of hydrogen $(2.02 u)$ is different from that of oxygen $(32.0 u)$.

We have:

$\frac{P V}{T}=0.26 J K^{-1}$

$R=8.314 J mole^{-1} K^{-1}$

Molecular mass $(M)$ of $H_2=2.02 u$

$\frac{P V}{T}=\mu R$ at constant temperature

Where, $\mu=\frac{m}{M}$

$m=$ Mass of $H_2$

$\therefore \quad m=\frac{P V}{T} \times \frac{M}{R}$

$=\frac{0.26 \times 2.02}{8.31}$

$=6.3 \times 10^{-2} g=6.3 \times 10^{-5} kg$

Hence, $6.3 \times 10^{-5} kg$ of $H_2$ will yield the same value of $P V / T$.

Answer

Volume of oxygen, $V_1=30$ litres $=30 \times 10^{-3} m^{3}$

Gauge pressure, $P_1=15 atm=15 \times 1.013 \times 10^{5} Pa$

Temperature, $T_1=27^{\circ} C=300 K$

Universal gas constant, $R=8.314 J mole^{-1} K^{-1}$

Let the initial number of moles of oxygen gas in the cylinder be $n_1$.

The gas equation is given as:

$P_1 V_1=n_1 R T_1$

$\therefore n_1=\frac{P_1 V_1}{R T_1}$

$=\frac{15.195 \times 10^{5} \times 30 \times 10^{-3}}{(8.314) \times 300}=18.276$

But, $n_1=\frac{m_1}{M}$

Where,

$m_1=$ Initial mass of oxygen

$M=$ Molecular mass of oxygen $=32 g$

$\therefore m_1=n_1 M=18.276 \times 32=584.84 g$

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, $V_2=30$ litres $=30 \times 10^{-3} m^{3}$

Gauge pressure, $P_2=11 atm=11 \times 1.013 \times 10^{5} Pa$

Temperature, $T_2=17^{\circ} C=290 K$

Let $n_2$ be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

$P_2 V_2=n_2 R T_2$

$\therefore n_2=\frac{P_2 V_2}{R T_2}$

$=\frac{11.143 \times 10^{5} \times 30 \times 10^{-3}}{8.314 \times 290}=13.86$

But, $n_2=\frac{m_2}{M}$

Where,

$m_2$ is the mass of oxygen remaining in the cylinder

$\therefore m_2=n_2 M=13.86 \times 32=453.1 g$

The mass of oxygen taken out of the cylinder is given by the relation:

Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder

$=m_1-m_2$

$=584.84 g-453.1 g$

$=131.74 g$

$=0.131 kg$

Therefore, $0.131 kg$ of oxygen is taken out of the cylinder.

Answer

Volume of the air bubble, $V_1=1.0 cm^{3}=1.0 \times 10^{-6} m^{3}$

Bubble rises to height, $d=40 m$

Temperature at a depth of $40 m, T_1=12^{\circ} C=285 K$

Temperature at the surface of the lake, $T_2=35^{\circ} C=308 K$

The pressure on the surface of the lake:

$P_2=1 atm=1 \times 1.013 \times 10^{5} Pa$

The pressure at the depth of $40 m$ :

$P_1=1 atm+d \rho g$

Where,

$\rho$ is the density of water $=10^{3} kg / m^{3}$

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

$\therefore P_1=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8=493300 Pa$

We have: $\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$

Where, $V_2$ is the volume of the air bubble when it reaches the surface

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

$=\frac{(493300)(1.0 \times 10^{-6}) 308}{285 \times 1.013 \times 10^{5}}$

$=5.263 \times 10^{-6} m^{3}$ or $5.263 cm^{3}$

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

Answer

Volume of the room, $V=25.0 m^{3}$

Temperature of the room, $T=27^{\circ} C=300 K$

Pressure in the room, $P=1 atm=1 \times 1.013 \times 10^{5} Pa$

The ideal gas equation relating pressure $(P)$, Volume $(V)$, and absolute temperature $(T)$ can be written as:

$P V=k_B N T$

Where,

$K_B$ is Boltzmann constant $=1.38 \times 10^{-23} m^{2} kg s^{-2} K^{-1}$

$N$ is the number of air molecules in the room

$ \begin{aligned} & \quad N=\frac{P V}{k_B T} \\ & =\frac{1.013 \times 10^{5} \times 25}{1.38 \times 10^{-23} \times 300}=6.11 \times 10^{26} \text{ molecules } \end{aligned} $

Therefore, the total number of air molecules in the given room is $6.11 \times 10^{26}$.

Answer

At room temperature, $T=27^{\circ} C=300 K$

Average thermal energy $=\frac{3}{2} k T$

Where $k$ is Boltzmann constant $=1.38 \times 10^{-23} m^{2} kg s^{-2} K^{-1}$

$\therefore \frac{3}{2} k T=\frac{3}{2} \times 1.38 \times 10^{-38} \times 300$

$=6.21 \times 10^{-21} J$

Hence, the average thermal energy of a helium atom at room temperature $(27^{\circ} C)$ is $6.21 \times$ $10^{-21} J$.

On the surface of the sun, $T=6000 K$

Average thermal energy $=\frac{3}{2} k T$

$=\frac{3}{2} \times 1.38 \times 10^{-38} \times 6000$

$=1.241 \times 10^{-19} J$

Hence, the average thermal energy of a helium atom on the surface of the sun is $1.241 \times$ $10^{-19} J$.

At temperature, $T=10^{7} K$

Average thermal energy $=\frac{3}{2} k T$

$=\frac{3}{2} \times 1.38 \times 10^{-23} \times 10^{7}$

$=2.07 \times 10^{-16} J$

Hence, the average thermal energy of a helium atom at the core of a star is $2.07 \times 10^{-16} J$.

Answer

Yes. All contain the same number of the respective molecules.

No. The root mean square speed of neon is the largest.

Since the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature.

According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, $N=6.023 \times 10^{23}$.

The root mean square speed ( $v_{rms}$ ) of a gas of mass $m$, and temperature $T$, is given by the relation:

$ v_{rms}=\sqrt{\frac{3 k T}{m}} $

Where, $k$ is Boltzmann constant

For the given gases, $k$ and $T$ are constants.

Hence $v_{\text{rms }}$ depends only on the mass of the atoms, i.e.,

$ v_{rms} \propto \sqrt{\frac{1}{m}} $

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.

Answer

Temperature of the helium atom, $T_{He}=-20^{\circ} C=253 K$

Atomic mass of argon, $M_{Ar}=39.9 u$

Atomic mass of helium, $M_{He}=4.0 u$

Let, $(v_{rms})_{Ar}$ be the rms speed of argon.

Let $(v_{rms})_{He}$ be the rms speed of helium.

The rms speed of argon is given by:

$(v_{rms})_{Ar} $

$=\sqrt{\frac{3 R T_{Ar}}{M_{Ar}}}\ldots(i)$

Where,

$R$ is the universal gas constant

$T_{Ar}$ is temperature of argon gas

The rms speed of helium is given by:

$(v_{rms})_{He}$

$=\sqrt{\frac{3 R T_{He}}{M_{He}}} \ldots($ ii $)$

It is given that:

$(v_{\text{rms }})_{Ar}$

$=(v_{rms})_{He}$

$ \begin{aligned} & \sqrt{\frac{3 R T_{Ar}}{M_{Ar}}}=\sqrt{\frac{3 R T_{He}}{M_{He}}} \\ \\ & \frac{T_{Ar}}{M_{Ar}}=\frac{T_{He}}{M_{He}} \\ \\ & T_{Ar}=\frac{T_{He}}{M_{He}} \times M_{Ar} \\ \\ & =\frac{253}{4} \times 39.9 \\ \\ & =2523.675=2.52 \times 10^{3} K \end{aligned} $

Therefore, the temperature of the argon atom is $2.52 \times 10^{3} K$.

Answer

Mean free path $=1.11 \times 10^{-7} m$

Collision frequency $=4.58 \times 10^{9} s^{-1}$

Successive collision time $\approx 500 \times($ Collision time $)$

Pressure inside the cylinder containing nitrogen, $P=2.0 atm=2.026 \times 10^{5} Pa$

Temperature inside the cylinder, $T=17^{\circ} C=290 K$

Radius of a nitrogen molecule, $r=1.0 \mathring{A}=1 \times 10^{10} m$

Diameter, $d=2 \times 1 \times 10^{10}=2 \times 10^{10} m$

Molecular mass of nitrogen, $M=28.0 g=28 \times 10^{-3} kg$

The root mean square speed of nitrogen is given by the relation: $v_{\text{rms }}=\sqrt{\frac{3 R T}{M}}$

Where,

$R$ is the universal gas constant $=8.314 J mole^{-1} K^{-1}$

$\therefore v_{\text{rms }}=\sqrt{\frac{3 \times 8.314 \times 290}{28 \times 10^{-3}}}=508.26 m / s$

The mean free path $(l)$ is given by the relation:

$l=\frac{k T}{\sqrt{2} \times d^{2} \times P}$

Where,

$k$ is the Boltzmann constant $=1.38 \times 10^{-23} kg m^{2} s^{-2} K^{-1}$

$\therefore l=\frac{1.38 \times 10^{-23} \times 290}{\sqrt{2} \times 3.14 \times(2 \times 10^{-10})^{2} \times 2.026 \times 10^{5}}$

$=1.11 \times 10^{-7} m$

Collision frequency $=\frac{v_{\text{rms }}}{l}$

$=\frac{508.26}{1.11 \times 10^{-7}}=4.58 \times 10^{9} s^{-1}$

Collision time is given as:

$T=\frac{d}{v_{\text{ms }}}$

$=\frac{2 \times 10^{-10}}{508.26}=3.93 \times 10^{-13} s$

Time taken between successive collisions:

$T^{\prime}=\frac{l}{v_{\text{ms }}}$

$ \begin{aligned} & =\frac{1.11 \times 10^{-7} m}{508.26 m / s}=2.18 \times 10^{-10} s \\ & \quad \frac{T^{\prime}}{T}=\frac{2.18 \times 10^{-10}}{3.93 \times 10^{-13}}=500 \end{aligned} $

Hence, the time taken between successive collisions is 500 times the time taken for a collision.