3 Marks Question - Physics STD 11 Science Questions - Vidyadip

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3 Marks Question

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Question 13 Marks

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m³ at a temperature of 27°C and 1 atm pressure.

Answer

Volume of the room, V = 25.0m³
Temperature of the room, T = 27°C = 300K
Pressure in the room, P = 1 atm = 1 × 1.013 × 10⁵Pa
The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as,
PV = k_BNT
Where,
K_B is Boltzmann constant = 1.38 × 10^-23m²kg s^-2k^-1
N is the number of air molecules in the room
$\therefore\text{ N}=\frac{\text{PV}}{\text{K}_{\text{B}}\text{T}}$
$=\frac{1.013\times10^{5}\times25}{(1.38\times10^{23}\times300)}$
= 6.11 × 10²⁶ molecules
Therefore, the total number of air molecules in the given room is 6.11 × 10²⁶

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Question 23 Marks

Molar volume is the volume occupied by 1mol of any (ideal) gas at standard temperature and pressure:
(STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.

Answer

The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:
PV = nRT
Where,
R is the universal gas constant = 8.314J mol^-1K^-1
n = Number of moles = 1
T = Standard temperature = 273K
P = Standard pressure = 1atm = 1.013 × 10⁵Nm^-2
$\therefore\ \text {V}=\frac{\text{nRT}}{\text{P}}$
$=\frac{1\times8.314\times273}{(1.013\times10^{5})}$
= 0.0224m³
= 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.

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Question 33 Marks

Uranium has two isotopes of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed ? If atomic mass of fluorine is 19 units, estimate the percentage difference in speeds at any temperature.

Answer

At a fixed temperature the average energy $=1 / 2 m\left\langle v^2\right\rangle$ is constant. So smaller the mass of the molecule, faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So
$
v_{349} / v_{352}=(352 / 349)^{1 / 2}=1.0044 .
$
Hence difference $\frac{\Delta V}{V}=0.44 \%$.
$\left[{ }^{235} U\right.$ is the isotope needed for nuclear fission. To separate it from the more abundant isotope ${ }^{238} U$, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and narrow, so that the molecule wanders through individually, colliding with the walls of the long pore. The faster molecule will leak out more than the slower one and so there is more of the lighter molecule (enrichment) outside the porous cylinder (Fig. 12.5). The method is not very efficient and has to be repeated several times for sufficient enrichment.].

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Question 43 Marks

A flask contains argon and chlorine in the ratio of $2: 1$ by mass. The temperature of the mixture is $27 C$. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed $V_{\text {rms }}$ of the molecules of the two gases. Atomic mass of argon $=39.9 u$; Molecular mass of chlorine $=70.9 u$.

Answer

The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to $(3 / 2) k_B T$. It depends only on temperature, and is independent of the nature of the gas.
(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is $1: 1$.
(ii) Now $1 / 2 m v_{ rms }^2=$ average kinetic energy per molecule $=(3 / 2)) k_{ B } T$ where $m$ is the mass of a molecule of the gas. Therefore,
$
\frac{\left( v _{r m s}^2\right)_{ Ar }}{\left( v _{r m s}^2\right)_{ Cl }}=\frac{(m)_{ Cl }}{(m)_{ Ar }}=\frac{(M)_{ Cl }}{(M)_{ Ar }}=\frac{70.9}{39.9}=1.77
$
where $M$ denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides,
$
\frac{\left( v _{r m s}\right)_{ Ar }}{\left( v _{r m s}\right)_{ Cl }}=1.33
$
You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered.

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Question 53 Marks

A vessel contains two nonreactive gases : neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2 u, molecular mass of O₂= 32.0 u.

Answer

Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since $V$ and $T$ are common to the two gases, we have $P_1 V=\mu_1 R T$ and $P_2 V=$ $\mu_2 R T$, i.e. $\left(P_1 / P_2\right)=\left(\mu_1 / \mu_2\right)$. Here 1 and 2 refer to neon and oxygen respectively. Since $\left(P_1 / P_2\right)=$ (3/2) (given), $\left(\mu_1 / \mu_2\right)=3 / 2$.
(i) By definition $\mu_1=\left(N_1 / N_{ A }\right)$ and $\mu_2=\left(N_2 / N_{ A }\right)$ where $N_1$ and $N_2$ are the number of molecules of 1 and 2 , and $N_{ A }$ is the Avogadro's number. Therefore, $\left(N_1 / N_2\right)=\left(\mu_1 / \mu_2\right)=3 / 2$.
(ii) We can also write $\mu_1=\left(m_1 / M_1\right)$ and $\mu_2=$ $\left(m_2 / M_2\right)$ where $m_1$ and $m_2$ are the masses of 1 and 2; and $M_1$ and $M_2$ are their molecular masses. (Both $m_1$ and $M_1$; as well as $m_2$ and $M_2$ should be expressed in the same units). If $\rho_1$ and $\rho_2$ are the mass densities of 1 and 2 respectively, we have
$
\begin{array}{l}
\frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left(\frac{M_1}{M_2}\right) \\
=\frac{3}{2} \times \frac{20.2}{32.0}=0.947
\end{array}
$

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Question 63 Marks

The density of water is 1000 $kg m ^{-3}$. The density of water vapour at $100^{\circ} C$ and $1 atm$ pressure is $0.6 kg m ^{-3}$. The volume of a molecule multiplied by the total number gives , what is called, molecular volume. The ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure is $6 \times 10^{-4}$. Estimate the volume of a water molecule using this data.

Answer

In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water $=1000 kg m ^{-3}$. To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to
$
(2+16) g =18 g =0.018 kg \text {. }
$
Since 1 mole contains about $6 \times 10^{23}$ molecules (Avogadro's number), the mass of a molecule of water is $(0.018) /\left(6 \times 10^{23}\right) kg =$ $3 \times 10^{-26} kg$. Therefore, a rough estimate of the volume of a water molecule is as follows :
Volume of a water molecule
$=\left(3 \times 10^{-26} kg \right) /\left(1000 kg m ^{-3}\right)$
$=3 \times 10^{-29} m ^3$
$=(4 / 3) \pi$ (Radius $^3$
Hence, Radius $\approx 2 \times 10^{-10} m =2 \mathring A$

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Question 73 Marks

At what temperature the mean speed of the molecules of hydrogen gas equals the escape epeed from the earth?

Answer

Mean speed of the molecule $=\sqrt{\frac{8\text{RT}}{\pi\text{M}}}$

Escape velocity $=\sqrt{2\text{gr}}$

$\sqrt{\frac{8\text{RT}}{\pi\text{M}}}=\sqrt{2\text{gr}}\Rightarrow\frac{8\text{RT}}{\pi\text{M}}=2\text{gr}$

$\Rightarrow\text{T}=\frac{2\text{gr}\pi\text{M}}{8\text{R}}=\frac{2\times9.8\times6400000\times3.14\times2\times10^{-3}}{8\times8.3}$ $=11863.9\approx11800\text{m/s}.$

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Question 83 Marks

Write ideal gas equation in terms of density.
If molar volume is the volume occupied by 1 mole of any (ideal) gas at STP, show that it is 22.4L (take R = 8.313 mol-¹ K^-1).

Answer

In terms of density, perfect gas equation is

$\text{p}=\frac{\rho\text{Rt}}{\text{M}_0}$

where, $\rho=$ mass density of the gas,

M₀ = molar mass of the gas, (mass of one mole of the gas)

R = 8.314J mol^-1K^-1

k_B = Boltmann's constant = 1.38 × 10^-23 J/K.

p = 1 atm = 0.76m of Hg

$=0.76\times(13.6\times10^3)\times9.8\text{Pa}$

$\Rightarrow\text{T}=273\text{K, R}=8.31\text{J mol}^{-1}\text{K}^{-1},\mu=1\text{mole}$

As $\text{pV}=\mu\text{RT}$

$\text{V}=\frac{\mu\text{RT}}{\text{p}}=\frac{1\times8.31\times273}{0.76\times(13.6\times10^3)\times9.8}$

$=22.4\times10^{-3}\text{m}^3=22.4\text{L}$

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Question 93 Marks

What is the relation between C_p and C_v?
Calculate the value of $\gamma$ (ratio between C_p and C_v) for diatomic gas.

Answer

C_p - C_v = R.
For diatomic gases, degree of freedom f = 5

$\therefore\text{C}_\text{v}=\frac{5}2{}\text{R}=\frac{5}{2}\times1.98$

$=4.95\text{cal/mole/k}$

$\therefore\text{C}_\text{p}=\Big(1+\frac{5}{2}\Big)\times1.98$

$=6.93\text{cal}/\text{mole}/\text{K}$

$\therefore\gamma=\frac{\text{C}_{\text{p}}}{\text{C}_{\text{v}}}=\frac{6.93}{4.95}=1.4$

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Question 103 Marks

On what parameter does the $\lambda$ (mean free path) depends?

Answer

We know that:

$\lambda=\frac{\text{kT}}{\sqrt{2}\pi\text{d}^2\rho}=\frac{\text{m}}{\sqrt{2}\pi\text{d}^2\rho}=\frac{1}{\sqrt{2}\pi\text{n}\text{d}^2}$

$\therefore\lambda$ depends upon:

Diameter (d) of the molecule, smaller the 'd', larger is the mean free path $\lambda.$
$\lambda\propto\text{T}$ i.e., higher the temperature larger is the $\lambda.$
$\lambda\propto\frac{1}{\text{P}}$ i.e., smaller the pressure larger is the $\lambda.$
$\lambda\propto\frac{1}{\text{P}}$ i.e., smaller the density $(\rho)$ larger will be the $\lambda.$
$\lambda\propto\frac{1}{\text{P}}$ i.e., smaller the number of molecules per unit volume of the gas, larger is the $\lambda.$

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Question 113 Marks

An ideal gas has a specific heat at constant pressure $\text{C}_{\text{p}}=\frac{5\text{R}}{2}.$ The gas is kept in a closed vessel of volume 0.0083m³ at a temperature of 300K and a pressure of 1.6 × 10⁶Nm^-2. An amount of 2.49 × 10⁴J of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.

Answer

P = 1.6 × 10⁶Nm^-2, V = 0.0083m³, T = 300K

We know that PV = nRT, where R = 8.3JK^-1 mol^-1

Therefore $\text{n}=\frac{\text{PV}}{\text{RT}}$

$=\frac{1.6\times10^6\times0.0083}{8.3\times300}=\frac{16}{3}\text{mole}$

Now $\text{C}_{\text{P}}-\text{C}_{\text{V}}=\text{R},$ therefore $\text{C}_{\text{V}}=\text{C}_{\text{P}}-\text{R}=\frac{5\text{R}}{2}-\text{R}=\frac{3\text{R}}{2}$

When heat energy Q is supplied to the gas, the inoease $\Delta\text{T}$ in its temperatue is obtained
from the relation

$\text{Q}=\text{nC}_{\text{V}}\Delta\text{T}$ or $\Delta\text{T}=\frac{\text{Q}}{\text{nC}_{\text{V}}}=\frac{2.49\times10^4}{\frac{16}{3}\times\frac{3}{2}\times8.3}=375\text{K}$

$\therefore$ Final temperature T’ = 300 + 375 = 675K. Since the gas is kept in a closed vessel, its volume remains constant. Hence the final pressure P is obtained from the relation.

$\frac{\text{P}'}{\text{T}'}=\frac{\text{P}}{\text{T}}$ or $\text{P}'=\text{P}\times\frac{\text{T}'}{\text{T}}$

$=\frac{1.6\times10^6\times675}{300}=3.6\times10^6\text{Nm}^{-2}$

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Question 123 Marks

If one mole of ideal monoatomic gas $\Big(\gamma=\frac{7}{5}\Big)$ is mixed with one mole of diatomic gas $\Big(\gamma=\frac{7}{5}\Big).$ What is the value of y for the mixtures?
(Here, $\gamma$ represents the ratio of specific heat at constant pressure to that at constant volume)

Answer

For monoatomic gas, $\text{C}_{\text{V}}=\frac{3}{2}\text{R}$

For diatomic gas, $\text{C}'_{\text{V}}=\frac{5}{2}\text{R}$

Let, $\mu$ and $\mu'$ be moles of mono and diatomic gases then. C_V (mixture) $=\frac{\mu\text{C}_{\text{V}}+\mu'\text{C}'_{\text{V}}}{\mu+\mu'}$

$\text{C}_{\text{V}}=\frac{1\times\frac{3}{2}\text{R}+1\times\frac{5}{2}\text{R}}{1+1}=2\text{R}$

$\gamma(\text{mixture})=1+\frac{\text{R}}{\text{C}_{\text{V (mixture)}}}$

$=1+\frac{\text{R}}{2\text{R}}=1.5$

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Question 133 Marks

What will be the mean free path of nitrogen gas at STP of given diameter of nitrogen molecule $=2\mathring{\text{A}}?$

Answer

Given, Diameter of nitrogen molecule, $\text{d}=2\mathring{\text{A}}$

$=2\times10^{-10}\text{m}$

At STP, one mole of gas (or 22.4L) of gas have

$\text{N}_{\text{A}}=6.023\times10^{23}\text{molecules}$

$\therefore$ Number density of nitrogen molecules

$\text{n}=\frac{\text{N}_{\text{A}}}{22.4\text{L}}=\frac{6.023\times10^{23}}{22.4\times10^{-3}\text{m}^3}$

$=2.69\times10^{25}\text{m}^{-3}$

$\therefore$ Mean free path of nitrogen at STP condition,

$\lambda=\frac{1}{\sqrt{2}\pi\text{nd}^2}$

$\lambda=\frac{1}{1.414\times3.142\times(2.69\times10^{25})\times(2\times10^{-10})^2}$

$=2.1\times10^{-7}\text{m}$

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Question 143 Marks

Figure. shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected into the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of 1 : 3. Find the ratio of the pressures in the two parts of the vessel.

Answer

$\text{n}_1=\text{n}_2=\text{n}$

$\text{P}_1=\frac{\text{nRT}}{\text{V}},\text{P}_2=\frac{\text{nRT}}{3\text{V}}$

$\frac{\text{P}_1}{\text{P}_2}=\frac{\text{nRT}}{\text{V}}\times\frac{3\text{V}}{\text{nRT}}=3:1$

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Question 153 Marks

State ideal gas equation. Draw graph to check whether a real gas obeys this equation. What is the conclusion drawn?

Answer

According to the ideal gas equation, we have PV $=\pi\text{RT}$

Thus, according to this equation $\frac{\text{PV}}{\pi\text{T}}=\text{R}$
i.e., value of $\frac{\text{PV}}{\pi\text{T}}$ must be a constant having a value 8.31J-mol^-1K^-1.
Experimentally value of $\frac{\text{PV}}{\pi\text{T}}$ for real gases was calculated by altering the pressure of gas at different temperatures. The graphs obtained have been shown in the figure. Here, for the purpose of comparison, graph for an ideal gas has also been drawn, which is a straight line parallel to pressure axis. From the graphs it is clear that behaviour of real gases differ from an ideal gas. However, at high temperatures and low pressures behaviour is nearly same as that of an ideal gas.

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Question 163 Marks

What do you mean by the rms speed of the molecules of the gas? Is rms speed same as the average speed at given temperature?

Answer

The square root of the mean of the squares of the random velocities of individual molecules of a gas is called rms speed of the molecules of the gas.

R.M.S. speed of the molecules of the gas.

$\text{r}_{\text{ms}}=\sqrt{\frac{\text{c}^2_1+\text{c}^2_2+...+\text{c}^2_{\text{n}}}{\text{n}}}=\sqrt{\frac{3\text{K}_{\text{B}}\text{T}}{\text{m}}}$

where n = number of molecules of the gas, m = mass of each molecule

K_B = Boltzmann constant, T = absolute temperature

Average speed is the speed with which a molecule of the gas moves.

$\text{c}_{\text{av}}=\frac{\text{c}_1+\text{c}_2+...+\text{c}_{\text{n}}}{\text{n}}=\sqrt{\frac{8\text{K}_{\text{B}}\text{T}}{\text{m}\pi}}$

where m is the mass of each molecule.

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Question 173 Marks

The volume of air bubble increases 15 times when it rises from bottom to the top of a lake. Calculate the depth of the lake if density of lake water is 1.02 × 10³kg/ m³ and atmospheric pressure is 75cm of mercury.

Answer

According to Boyle's law,

P₁V₁ = P₂V₂ ...(1)

Here P₁ = 75cm of Hg = 75m of Hg = 0.75 × 13.6 × 10³ × 9.8

= 99.96 × 10³Nm^-2

Let volume of bubble at depth h = x

i.e., V₂ = x $\therefore$ V₁ = 16x

P₂ = 75cm of Hg + hp_water g = 99.96 × 10³ + h × 10³× 9.8

Using eqn. (1), we get

99.96 × 10³ × 16x = (99.96 × 10³ + h × 10³ × 9.8)x

9.8h = 99.96 × 16 - 99.96 = 99.96 × 15

$\therefore\text{h}=\frac{99.96\times15}{9.8}=153\text{m}$

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Question 183 Marks

Two perfect gases at absolute temperatures T₁ and T₂ are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m₁ and m₂ and the number of the molecules in the gases are n₁ and n₂ respectively.

Answer

According to kinetic theory, the average kinetic energy per molecule of a gas $=\frac{3}{2}\text{k}_{\text{B}}\text{T}$

Before mixing, the two gases, the average K.E. of all the molecules of two gases

$=\frac{3}{2}\text{k}_{\text{B}}\text{n}_1\text{T}_1+\frac{3}{2}\text{k}_{\text{B}}\text{n}_2\text{T}_2$

After mixing, the average K.E. of both the gases $=\frac{3}2{}\text{k}_{\text{B}}(\text{n}_1+\text{n}_2)\text{T}$

where T is the temperature of mixture. Since there is no loss of energy,

Hence, $\frac{3}{2}\text{k}_{\text{B}}(\text{n}_1+\text{n}_2)\text{T}$

$=\frac{3}{2}\text{k}_{\text{B}}\text{n}_1\text{T}_1+\frac{3}2{}\text{k}_{\text{B}}\text{n}_2\text{T}_2$

$\text{T}=\frac{\text{n}_1\text{T}_1+\text{n}_2\text{T}_2}{(\text{n}_1+\text{n}_2)}$

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Question 193 Marks

An electric bulb of volume 250cm³ was sealed off during manufacture at a pressure of 10^-3 mm of Hg at 27°C. Find the number of air molecules in the bulb.

Answer

Let N be the number of air molecules in the bulb. It is given that

P = 10^-3mm of Hg = 10^-4cm of Hg

V = 250cm³

T = 273 + 27 = 300K

Now PV = NkT

We know that at STP, one mole of a gas occupies a volume V₀ = 22400cm³ and contains N₀ = 6.02 × 10²³ molecules (N₀ is the Avogadro's number) and the pressure P₀ = 76cm of Hg and T₀ = 273K. Also

P₀V₀ = N₀ KT₀ …(ii)

Dividing eqn. (i) and (ii), we get

$\text{N = N}_0\times\frac{\text{T}_0}{\text{T}}\times\frac{\text{P}}{\text{P}_0}\times\frac{\text{V}}{\text{V}_0}$

$=(6.02\times10^{23})\times\Big(\frac{273}{300}\Big)\\\times\Big(\frac{10^{-4}}{76}\Big)\times\Big(\frac{250}{22400}\Big)$

$=8.045\times10^{15}\text{molecules.}$

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Question 203 Marks

Figure. shows a vessel partitioned by a fixed diathernnc separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part to the mass of a molecule in the right part.

Answer

The left side of the container has a gas, let having molecular wt. M₁

Right part has Mol. wt = M₂

Temperature of both left and right chambers are equal as the separating wall is diathermic,

$\sqrt{\frac{3\text{RT}}{\text{M}_1}}=\sqrt{\frac{8\text{RT}}{\pi\text{M}_2}}\Rightarrow\frac{3\text{RT}}{\text{M}_1}=\frac{8\text{RT}}{\pi\text{M}_2}$

$\Rightarrow\frac{\text{M}_1}{\pi\text{M}_2}=\frac{3}{8}\Rightarrow\frac{\text{M}_1}{\text{M}_2}=\frac{3\pi}{8}=1.1778\approx1.18$

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Question 213 Marks

You have the following group of particles, n_irepresents no. of molecules with speed v_i(ms^-1):

n_i	2	4	8	6	3
v_i	1.0	2.0	3.0	4.0	5.0

Calculate:

The average speed.
The r.m.s. speed.
The most probable speed.

Answer

n_i = Number of molecules

v_i = Speed of molecules

n_i	2	4	8	6	3
v_i	1.0	2.0	3.0	4.0	5.0

The average speed

$\text{v}_{\text{avg}}=\frac{\text{n}_1\text{v}_1+\text{n}_2\text{v}_2+\text{n}_3\text{v}_3+...+\text{n}_5\text{v}_5}{\text{n}_1+\text{n}_2+\text{n}_3+...+\text{n}_5}$

$=\frac{(2\times1)+(4\times2)+(8\times3)+(6\times4)+(3\times5)}{2+4+8+6+3}$

$=\frac{2+8+24+24+15}{23}=\frac{73}{23}=3.17\text{ms}^{-1}$

$\text{v}_{\text{rms}}=\sqrt{\frac{\text{n}_1\text{v}_1^2+\text{n}_2\text{v}_2^2+\text{n}_3\text{v}_3^2+\text{n}_4\text{v}_4^2+\text{n}_5\text{v}_5^2}{\text{n}_1+\text{n}_2+\text{n}_3+\text{n}_4+\text{n}_5}}$

$=\sqrt{\frac{2(1)^2+4(2)^2+8(3)^2+6(4)^2+3(5)^2}{2+4+8+6+3}}$

$=\sqrt{\frac{2+16+72+96+75}{23}}$

$=\sqrt{\frac{261}{23}}=\sqrt{11.347}$

$=3.37\text{ms}^{-1}$

The most probable sped $=3\text{ms}^{-1}$

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Question 223 Marks

Explain the pressure exerted by an ideal gas and also find the average kinetic energy per molecule of the gas.

Answer

From kinetic theory of gases, the pressure P exerted by an ideal gas of density p and r.m.s. velocity of its gas molecules C is given by

$\text{P}=\frac{1}{3}\rho\text{C}^2$

Mass of unit volume of the gas $=1\times\rho=\rho$

Mean kinetic energy of translation per unit volume of the gas is

$\text{E}=\frac{1}{2}\rho\text{C}^2,$

$\therefore\frac{\text{P}}{\text{E}}=\frac{\big(\frac{1}{3}\big)\rho\text{C}^2}{\big(\frac{1}{2}\big)\rho\text{C}^2}=\frac{2}{3}$

$\text{P}=\frac{2}{3}\text{E}$

"The pressure exerted by an ideal gas is numerically equal to two third of the mean kinetic energy of translation per unit volume of the gas."

Average Kinetic Energy per Molecule of the Gas: Consider one gram mole of an ideal gas occupying a volume V at temperature T. Let m be the mass of each molecule of the gas. Then

M = m × N_A

where N_A is Avogadro's number.

If C is the r.m.s. velocity of the gas molecules, then pressure P exerted by ideal gas is

$\text{P}=\frac{1}{3}\rho\text{C}^2=\frac{1}{3}\frac{\text{M}}{\text{V}}\rho\text{C}^2$

$\text{PV}=\frac{1}{3}\text{MC}^2$

From perfect gas equation, PV = RT, where R is a universal gas constant for one gram mole of the gas.

$\therefore\frac{1}{3}\text{MC}^2=\text{RT}$

$\frac{1}{3}\text{MC}^2=\frac{3}{2}\text{RT}$

$\therefore$ Average kinetic energy of translation of one mole of the gas $=\frac{1}{2}\text{MC}^2=\frac{3}{2}\text{RT}$

$\frac{1}{2}\text{mN}_{\text{A}}\text{C}^2=\frac{3}{2}\text{RT}$ $(\because\text{M = mN}_{\text{A}})$

$=\frac{1}{2}\text{mC}^2=\frac{3}{2}\Big(\frac{\text{R}}{\text{N}_{\text{A}}}\Big)\text{T}=\frac{3}{2}\text{k}_{\text{B}}\text{T}$ $\Big(\because\frac{\text{R}}{\text{N}}=\text{k}_{\text{B}}\Big)$

where k_B is called Boltzmarm constant.

$\therefore$ Average K.E. of trarulation per molecule of gas $=\frac{1}{2}\text{mC}^2=\frac{3}{2}\text{k}_{\text{B}}\text{T}.$

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Question 233 Marks

What will be the internal energy of 8g of oxygen at STP?

Answer

Oxygen is a diatomic gas

Number of moles of O₂ gas

$=\frac{\text{Atomic wt.}}{\text{Molecular wt.}}=\frac{8}{32}$

$=\frac{1}{4}=0.25$

$\therefore$ Energy associated with 1 mole of oxygen

$\text{U}=\frac{5}2{}\text{RT}$

$\therefore$ Internal energy of 8g of oxygen $=0.25\times\frac{5}{2}\text{RT}$

$=0.25\times\frac{5}{2}\times8.31\times273=1417.9\text{J}$

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Question 243 Marks

The average translational kinetic energy of air molecules is 0.040eV (1eV = 1.6 × 10^-19J). Calculate the temperature of the air. Boltzmann constant k = 1.38 × 10^-23JK^-1.

Answer

Agv. K.E. $=\frac{3}{2}\text{KT}$

$\Rightarrow \frac{3}{2}\text{KT}=0.04\times1.6\times10^{-19}$

$\Rightarrow\Big(\frac{3}{2}\Big)\times1.38\times10^{-23}\times\text{T}=0.04\times1.6\times10^{-19}$

$\Rightarrow\text{T}=\frac{2\times0.04\times1.6\times10^{-19}}{3\times1.38\times10^{-23}}$ $=0.0309178\times10^4=309.178\approx310\text{K}$

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Question 253 Marks

Calculate the temperature atoms at which rms speed of Argon gas is equal to the rms speed of Helium gas atoms at -10°C? (Atomic mass of Ar = 39.9u, that of He = 4u)

Answer

As we know that, V_rms $=\sqrt{\frac{3\text{RT}}{\text{M}}}$

Thus, $\frac{\text{V}_{\text{rms}}}{\text{AR}}=\frac{\text{V}_{\text{rms}}}{\text{He}}$

$\Rightarrow\sqrt{\frac{\text{T}_{\text{Ar}}}{\text{M}_{\text{Ar}}}}=\sqrt{\frac{\text{T}_{\text{He}}}{\text{M}_{\text{He}}}}$

$\text{T}_{\text{Ar}}=?\text{ T}_{\text{He}}=273-10=263\text{K}$

$\text{M}_{\text{Ar}}=39.9\text{u, M}_{\text{He}}=4\text{u}$

Thus, $\frac{\text{T}_{\text{Ar}}}{39.9}=\frac{263}{4}$

$\text{T}_{\text{Ar}}=\frac{263\times39.9}{4}=2623.43\text{K}$

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Question 263 Marks

The molecules of a given mass of a gas have root mean square speeds of 100m/s at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

Answer

Here $\text{C}_1=100\text{m/s C}_2=?$

$\text{T}_1=27^{\circ}\text{C}=(27+273)\text{K}=300\text{K}$

$\text{T}_2=127^{\circ}\text{C}=(127+273)\text{K}=400\text{K}$

$\text{P}_1=1.00\text{ atm}$

$\text{P}_2=2.0\text{ atm}$

From $\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2};\frac{\text{V}_1}{\text{V}_2}=\frac{\text{P}_2}{\text{P}_1}\frac{\text{T}_1}{\text{T}_2}$

$=2\times\frac{300}{400}=\frac{3}{2}$

Again $\text{P}_1=\frac{1}{3}\frac{\text{M}}{\text{V}_1}\text{C}^2_1$ and $\text{P}_2=\frac{1}{3}\text{M}$

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Question 273 Marks

Give a formula for mean free path of the molecules of a gas. Briefly explain, how its value is affected by (i) change in temperature and (ii) change in pressure.

Answer

As, we know that the value of mean free path of the molecules of a given gas is given by

Mean free path, $\lambda=\frac{1}{\sqrt{2}\pi\text{nd}^2}$

Here, n = number of gas molecules present in unit volume of given gas and d = molecular diameter.

Effect of temperature: As temperature of a gas is increased at constant pressure, volume of gas increases and hence n, the number of molecules per unit volume decrease. In fact

$\text{n}\propto\frac{1}{\text{V}}$ and $\text{V}\propto\text{T},$

Thus, $\text{n}\propto\frac{1}{\text{T}}$

Due to decrease in molecular number density, the value of mean free path of the gas increase i.e. $\lambda\propto\frac{1}{\text{n}}\propto\text{T}.$ Thus, pressure remaining constant, the mean free path of a gas is directly proportional to its absolute temperature.

Effect of pressure: At constant temperature, on increasing pressure, the volume V decrease, the molecular number density n increases and consequently, the mean free path decreases

i.e. $\text{p}\propto\frac{1}{\text{V}}\propto\text{n}$

$\therefore\lambda\propto\frac{1}{\text{n}}$

$\lambda\propto\frac{1}{\text{p}}$

Thus, at a constant temperature, the mean free path of a gas is inversely proportional to its pressure.

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Question 283 Marks

Calculate the total number of degrees of freedom possessed by the molecules in one cm³ of H₂ gas at NTP.

Answer

22400cm³ of every gas contains 6.02 × 10²³ molecules.

$\therefore$ Number of molecules in 1cm³ of H₂ gas

$=\frac{6.02\times10^{23}}{22400}=0.26875\times10^{20}$

Number of degrees of freedom of a H₂ gas molecule = 5

$\therefore$ Total number of degrees of freedom of 0.26875 × 10²⁰ molecules

= 0.26875 ×10²⁰ × 5 = 1.34375 × 10²⁰

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Question 293 Marks

Two ideal monoatomic gases A and B at 27°C and 37°C are mixed. The number of moles in gas A are 2 and number of moles in gas B are 3. What will be the temperature of the mixture?

Answer

Sum of K.E. of gases A and B = K.E. of the mixture

$\mu_1\Big(\frac{3}{2}\text{RT}_1\Big)+\mu_2\Big(\frac{3}2{}\text{RT}_2\Big)$

$=(\mu_1+\mu_2)\Big(\frac{3}{2}\text{RT}\Big)$

$\therefore\text{T}=\frac{\mu_1\text{T}_1+\mu_2\text{T}_2}{\mu_1+\mu_2}$

$=\frac{2\times300+3\times310}{2+3}=306\text{K}$

$\therefore$ Temperature of mixture $=33^{\circ}\text{C}$

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Question 303 Marks

An electric bulb of volume 250cc was sealed during manufacturing at a pressure of 10^-3mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 × 10²³mol^-1, density of mercury = 13600kg/ m^-3 and g = 10ms^-2.

Answer

V = 250cc = 250 × 10^-3

P = 10^-3mm = 10^-3 × 10^-3m

= 10^-6 × 13600 × 10 pascal

= 136 × 10^-3 pascal

T = 27°C = 300K

$\text{n}=\frac{\text{PV}}{\text{RT}}=\frac{136\times10^{-3}\times250}{8.3\times300}\times10^{-3}$

$=\frac{136\times250}{8.3\times300}\times10^{-6}$

No. of molecules $=\frac{136\times250}{8.3\times300}\times10^{-6}\times6\times10^{23}$

$=81\times10^{17}\approx0.8\times10^{15}$

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Question 313 Marks

Two moles of gas A at 27°C are mixed with 3 moles of gas B at 37°C. If both are monoatomic ideal gases, what will be the temperature of the mixture?

Answer

As there is no loss of energy in the process, therefore, sum of KE of gases A and B = KE of mixture,

$\mu_1\Big(\frac{3}{2}\text{RT}_1\Big)+\mu_2\Big(\frac{3}{2}\text{RT}_2\Big)$

$=(\mu_1+\mu_2)\frac{3}{2}\text{RT}$

where T is temperature of the mixture.

$\therefore\text{T}=\frac{\mu_1\text{T}_1+\mu_2\text{T}_2}{\mu_1+\mu_2}$

$=\frac{2(27+273)+3(37+273)}{2+3}$

$=\frac{600+930}{5}=\frac{1530}{5}$

$=306-273=3^{\circ}\text{C}$

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Question 323 Marks

Using the expression for pressure exerted by a gas, deduce Avogadro's law and Graham's law of diffusion.

Answer

Avogadro's law: Consider two gases A and B having equal volume V at same temperature T and pressure P.

For gas A,

$\text{P}=\frac{1}{3}\frac{\text{M}\rho_1}{\text{V}}\text{c}^2_1$

For gas B,

$\text{P}=\frac{1}{3}\frac{\text{M}_2}{\text{V}}\text{c}^2_2$

$\therefore\text{P}=\frac{\text{M}_1\text{c}^2_1}{3\text{V}}=\frac{\text{M}_2\text{c}^2_2}{3\text{V}}$

or $\text{M}_1\text{c}^2_1=\text{M}_2\text{c}^2_2 \ ...(\text{i})$

As gases are at the same temperature.

Average K.E. per molecule of gas A = Average K.E. per molecule of gas A

$\frac{1}{2}\text{m}_1\text{c}^2_1=\frac{1}{2}\text{m}_2\text{c}^2_2$

$\text{m}_1\text{c}^2_1=\text{m}_2\text{c}^2_2 \ ...(\text{ii})$

Dividing (i) by (ii)

$\frac{\text{M}_1\text{c}^2_1}{\text{m}_1\text{c}^2_1}=\frac{\text{M}_2\text{c}^2_2}{\text{m}_2\text{c}^2_2}$

$\frac{\text{m}_1\text{n}_1\text{c}^2_1}{\text{m}_1\text{c}^2_1}=\frac{\text{m}_2\text{n}_2\text{c}^2_2}{\text{m}_2\text{c}^2_2}$

$\therefore\text{n}_1=\text{n}_2$ (which is Avogadro's law)

Graham's law:

$\text{P}_1=\frac{1}{3}\rho_1\text{c}^2_1\text{P}_2=\frac{1}{3}\rho_2\text{c}^2_2$

When the steady state of diffusion is reached,

$\text{P}_1=\text{P}_2$

$\frac{1}{3}\rho_1\text{c}^2_1=\frac{1}{3}\rho_2\text{c}^2_2$

$\frac{\text{c}^2_1}{\text{c}^2_2}=\frac{\rho_2}{\rho_1}$

or $\frac{\text{c}_1}{\text{c}_2}=\sqrt{\frac{\rho_2}{\rho_1}}$

If r₁ and r₂ are rates of diffusion of A and B respectively,

then

$\text{r}_1\propto\text{c}_1,\text{r}_2\propto\text{C}_2$

$\frac{\text{r}_1}{\text{r}_2}=\frac{\text{c}_1}{\text{c}_2}$

or $\frac{\text{r}_1}{\text{r}_2}=\sqrt{\frac{\rho_2}{\rho_1}}$

$\Rightarrow\text{r}\propto\frac{1}{\sqrt{\rho}}$

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Question 333 Marks

At what temperature will the average velocity of oxygen molecules be sufficient so as to escape from the earth? Escape velocity from the earth is 11.0km/sec and the mass of one molecule of oxygen is 5.34 × 10^-26kg (Boltzmann constant k = 1.38 × 10^-23 Joule/K).

Answer

We know $\frac{1}2{}\text{mv}^2=\frac{3}{2}\text{kT}$ or $\text{T}=\frac{\text{mv}^2}{3\text{k}}$

Here, $\text{m}=5.34\times10^{-26}\text{kg}$

$\text{v}=11.0\text{km/s}=11\times10^3\text{ms}^{-1}$

$\text{k}=1.38\times10^{-23}\text{Jk}^{-1}$

$\therefore\text{T}=\frac{5.34\times10^{-26}\times(11\times10^3)^2}{3\times1.38\times10^{-23}}$

$=1.56\times10^5\text{K}$

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Question 343 Marks

An enclosure of volume four litres contains a mixture of 8g of oxygen, 14g of nitrogen and 22g of carbon dioxide. If the temperature of the mixture is 27°C, find the pressure of the mixture of gases. Given R = 8.315J-K^-1mol^-1.

Answer

Temperature T = 300K

Volume V = 4 liter = 4 × 10^-3m³

The pressure exerted by a gas is given by

$\text{P}=\frac{\text{nRT}}{\text{V}}=\frac{\text{mass}}{\text{molecular weight}}\times\frac{\text{RT}}{\text{V}}$

Pressure exerted by oxygen $\text{P}_1=\frac{8}{32}\frac{\text{RT}}{\text{V}}=\frac{1}{4}\frac{\text{RT}}{\text{V}}$

Pressure exerted by nitrogen $\text{P}_2=\frac{14}{28}\frac{\text{RT}}{\text{V}}=\frac{1}{2}\frac{\text{RT}}{\text{V}}$

Pressure exerted by carbon dioxide $\text{P}_3=\frac{22}{44}\frac{\text{RT}}{\text{V}}=\frac{1}2{}\frac{\text{RT}}{\text{V}}$

From Dalton's law of partial pressures, the total pressure exerted by the mixture is given by

$\text{P = P}_1+\text{P}_2+\text{P}_3$

$=\frac{1}{4}\frac{\text{RT}}{\text{V}}+\frac{1}{2}\frac{\text{RT}}{\text{V}}+\frac{1}{2}\frac{\text{RT}}{\text{V}}$

$=\frac{5}{4}\frac{\text{RT}}{\text{V}}=\frac{\text{RT}}{\text{V}}=\frac{5}{4}\times\frac{8.315\times300}{4\times10^{-3}}$

$=7.79\times10^5\text{Nm}^{-2}$

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Question 353 Marks

A gas cylinder has walls that can bear a maximum pressure of 1.0 × 10⁶Pa. It contains a gas at 8.0 × 10⁵Pa and 300K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

Answer

P₁ = 8.0 × 10⁵ Pa, P₂ = 1 × 10⁶Pa, T₁ = 300K, T₂ = ?

Since, V₁ = V₂ = V

$\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}$

$\Rightarrow\frac{8\times10^5\times\text{V}}{300}=\frac{1\times10^6\times\text{V}}{\text{T}_2}$

$\Rightarrow\text{T}_2=\frac{1\times10^6\times300}{8\times10^5}=375^\circ\text{K}$

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Question 363 Marks

Show that the average K.E. of a gas molecule is directly proportional to the temperature of the gas. Hence give the kinetic interpretation of temperature.

Answer

Consider 1g mole of an ideal gas occupying a volume V at temperature T.

Let m be the mass of each molecule of the gas. Then M = m × N_A.

where N_A is Avogadro's number.

If C is the r.m.s. velocity of the gas molecules, then pressure P exerted by ideal gas is

$\text{MC}^2\text{ P}=\frac{1}{3}\rho\text{C}^2=\frac{1}{3}\frac{\text{M}}{\text{V}}\text{C}^2$ or $\text{PV}=\frac{1}{3}$

From perfect gas equation, PV = RT, where R is a universal gas constant for 1g mole of the gas.

$\therefore\frac{1}{3}\text{MC}^2=\text{RT}$

$\frac{1}{3}\text{MC}^2=\frac{3}{2}\text{RT}$

$\therefore$ Average of K.E. of translation of 1 mole of the gas

$=\frac{1}{2}\text{MC}^2$

$\Rightarrow\text{E}=\frac{3}{2}\text{RT}$

According to kinetic theory of gases, the pressure P exerted by 1 mole of an ideal gas is given by

$\text{P}=\frac{1}{3}\rho\text{C}^2=\frac{1}{3}\frac{\text{M}}{\text{V}}\text{C}^2$

$\text{PV}=\frac{1}{3}\text{MC}^2$

$\frac{1}3{}\text{MC}^2=\text{RT}$ $(\because\text{PV = RT})$

$\Rightarrow\text{C}^2=\frac{3\text{RT}}{\text{M}}\Rightarrow\text{C}^2\propto\text{T}$ ($\because$ R and M are constants)

$\Rightarrow\text{C}^2\propto\sqrt{\text{T}}\Rightarrow\sqrt{\text{T}}\propto\text{C}$

Thus, the square root of the absolute temperature of an ideal gas is directly proportional to root mean square velocity of its molecules.

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Question 373 Marks

A cylinder of fixed capacity contains 44.8L of helium gas at STP. Calculate the amount of heat required to raise the temperature of container by 15°C? [given R = 8.31J-mol^-1K^-1]

Answer

At STP, 1 mole of gas occupy 22.4L of volume.

$\therefore$ Moles of helium in container, $\mu=\frac{44.8}{22.4}=2\text{ moles}$

Now, helium is monoatomic, so, $\text{C}_{\text{V}}=\frac{3}{2}\text{R}$

Change in temperatue,

$\Delta\text{T = T}_2-\text{T}_1=15^{\circ}\text{C}=15\text{K}$

$\therefore\Delta\text{W}=\text{p}\Delta\text{V}=0$

$\Rightarrow\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$

Amourt of heat required, $\Delta\text{Q}=\Delta\text{U}=\mu\text{C}_{\text{V}}\Delta\text{T}$

$=2\times\frac{3}{2}\text{R}\times15=45\text{R}$

$=45\times8.31=374\text{J}$

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Question 383 Marks

2g of hydrogen is sealed in a vessel of volume 0.02m³ and is maintained at 300K. Calculate the pressure in the vessel.

Answer

m = 2g, V = 0.02m³ = 0.02 × 10⁶cc = 0.02 × 10³L, T = 300K, P = ?

M = 2g,

$\text{PV}=\text{nRT}\Rightarrow\text{PV}=\frac{\text{m}}{\text{M}}\text{RT}$

$\Rightarrow\text{P}\times20=\frac{2}{2}\times0.082\times300$

$\Rightarrow\text{P}=\frac{0.082\times300}{20}$

$=1.23\text{ atm}=1.23\times10^5\text{ pa}\approx1.23\times10^5\text{pa}$

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Question 393 Marks

The temperature and relative humidity in a room are 300K and 20% respectively. The volume of the room is 50m³. The saturation vapour pressure at 300K is 3.3kPa. Calculate the mass of the water vapour present in the room.

Answer

T = 300K, Rel. humidity = 20%, V = 50m³

SVP at 300K = 3.3KPa, V.P. = Relative humidity × SVP = 0.2 × 3.3 × 10³

$\text{PV}=\frac{\text{m}}{\text{M}}\text{RT}$

$\Rightarrow0.2\times3.3\times10^3\times50=\frac{\text{m}}{18}\times8.3\times300$

$\Rightarrow\text{m}=\frac{0.2\times3.3\times50\times18\times10^3}{8.3\times300}=238.55\text{g}\approx238\text{g}$

Mass of water present in the room = 238g.

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Question 403 Marks

Calculate the temperature at which r.m.s. velocity of a gas molecule is same as that of a molecule of another gas at 47°C. Molecular weight of first and second gases are 64 and 32 respectively.

Answer

As $\text{P}=\frac{1}{3}\frac{\text{M}_{\text{m}}}{\text{V}}\text{v}^2_{\text{rms}}$ or $\text{PV}=\frac{1}{2}\text{M}_{\text{m}}\text{v}^2_{\text{rms}}$

or $\text{RT}=\frac{1}3{}\text{M}_{\text{m}}\text{v}^2_{\text{rms}}$ or $\frac{\text{M}_{\text{m}}\text{v}^2_{\text{rms}}}{\text{T}}=3\text{R}=\text{Constant}$

If $(\text{M}_{\text{m}})_1$ and $(\text{M}_{\text{m}})_2$ be the molecular weights of the two gases, $\text{v}_{\text{rms}}=\text{C}^2_1\text{ and }\text{C}^2_2,$ the mean square velocities of two gases and T₁ and T₂ are the absolute temperatures of the two gases, then

$\frac{(\text{M}_{\text{m}})_1\text{C}^2_1}{\text{T}_1}=\frac{(\text{M}_{\text{m}})_2\text{C}^2_2}{\text{T}_2}$

According to the given problem

$\sqrt{\text{C}^2_1}=\sqrt{\text{C}^2_2}$ or $\text{C}^2_1=\text{C}^2_2$

$\therefore\frac{(\text{M}_{\text{m}})_1}{\text{T}_1}=\frac{(\text{M}_\text{m})_2}{\text{T}_2}$ or $\frac{64}{\text{T}_1}=\frac{32}{320}$ $[\because\text{T}_2=47^{\circ}\text{C}=320\text{K}]$

$\text{T}_1=640^{\circ}=267^{\circ}\text{C}$

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Question 413 Marks

Calculate the temperature at which the root mean square velocity of nitrogen molecules will be equal to 8km-s^-1.

Answer

Given, r.m.s. velocity, C = 8km-s^-1 = 8 × 10⁵cm-s^-1

Molar gas constant = R = 8.31 × 10⁷erg mol^-1K^-1

Molecular weight of nitrogeru M = 28

Let T be the required temperature

Then, $\text{C}=\sqrt{\frac{3\text{RT}}{\text{M}}}$ or $\text{C}^2=\frac{3\text{RT}}{\text{M}}$ or $\text{T}=\frac{\text{MC}^2}{3\text{R}}$

$=\frac{28\times(8\times10^5)^2}{3\times8.31\times10^7}\text{K}=71881\text{K}$

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Question 423 Marks

At what temperature would the root-mean square speed of a gas molecule have twice its value at 100°C?

Answer

We know that

$\text{C}^2=3\frac{\text{RT}}{\text{Nm}}=3\frac{\text{Kt}}{\text{m}}$

Thus $\text{C}^2_1=\frac{3\text{kT}_1}{\text{m}}$ and $\text{C}_2^2=\frac{3\text{kT}_2}{\text{m}}$

$\therefore\frac{\text{C}^2_1}{\text{C}^2_2}=\frac{\text{T}_1}{\text{T}_2}$

Here $\text{C}_2=2\text{C}_1,\text{T}=273+100=373\text{K}$

$\text{T}_2=\text{T}_1\times\frac{\text{C}^2_2}{\text{C}^2_1}$

$=373\times4=1492\text{K}=1219^{\circ}\text{C}$

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Question 433 Marks

Equal masses of air are sealed in two vessels, one of volume V₀ and the other of volume 2V₀. If the first vessel is maintained at a temperature 300K and the other at 600K, find the ratio of the pressures in the two vessels.

Answer

Since mass is same

n₁ = n₂ = n

$\text{P}_1=\frac{\text{nR}\times300}{\text{V}_0},\text{P}_2=\frac{\text{nR}\times600}{2\text{V}_0}$

$\frac{\text{P}_1}{\text{P}_2}=\frac{\text{nR}\times300}{\text{V}_0}\times\frac{2\text{V}_0}{\text{n}\text{R}\times600}=\frac{1}{1}=1:1$

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Question 443 Marks

Three moles of an ideal diatomic gas is taken at a temperature of 300K. Its volume is doubled keeping its pressure constant. Find the change in internal energy of gas.

Answer

Here, $\mu=3,\text{T}_1=300\text{K}$ and for an ideal monoatomic gas

$\text{C}_{\text{v}}=\frac{5}{2}\text{R}$

As volume of gas is doubled (V₂ = 2V₁) at corstant pressure, hence according to Charlds law

$\text{T}_2=\frac{\text{T}_1\text{V}_2}{\text{V}_1}=\frac{300\times2\text{V}_1}{\text{V}_1}=600\text{K}$

$\therefore$ Gain in internal energy $\text{u}_2-\text{u}_1=\mu\text{C}_{\text{v}}(\text{T}_2-\text{T}_1)$

$=3\times\frac{5}{2}\text{R}\times(600-300)$

$=2250\text{R}=2250\times8.31\times10^4\text{J}$

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Question 453 Marks

Give the essential feature of kinetic theory of gases. Show that the pressure exerted by a gas is equal to two-third of the average kinetic energy.

Answer

Postulates of kinetic theory of gases:

A gas consists of a very large number of molecules (of the order of Avogadro's number, 10²³), which are perfect elastic spheres. They are identical in all respects for a given gas and are different for different gases.
The molecules of a gas are in a state of incessant random motion. They move in all directions with different speeds and obey Newton's laws of motion.
The molecules do not exert any force of attraction or repulsion on each other, except during collision.

From the kinetic theory of gases, the pressure P exerted by an ideal gas of density $\rho$ and r.m.s. velocity of its gas molecules c is given by

$\text{p}=\frac{1}{3}\rho\text{c}^2$

Mass of unit volume of the gas $=1\times\rho=\text{p}$

Mean kinetic energy of translation per unit volume of the gas is

$\text{E}=\frac{1}{2}\rho\text{c}^2$

$\therefore\frac{\text{P}}{\text{E}}=\frac{\big(\frac{1}{3}\big)\rho\text{c}^2}{\big(\frac{1}{2}\big)\rho\text{c}^2}=\frac{2}{3}\Rightarrow\text{P}=\frac{2}{3}\text{E}$

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Question 463 Marks

Calculate the root-mean square speed of oxygen molecules at 1092K. Density of oxygen at STP = 1.424kg-m^-3.

Answer

We first calculate the root-mean square speed of oxygen at STP.

$\text{P}_0=0.76\text{m}$ of $\text{Hg}=1.01\times10^5\text{Nm}^{-2}$

$\rho_0=1.424\text{kg-m}^{-3}$

The root-mean square speed at 0°C is given by

$\text{c}_0=\sqrt{\frac{3\text{P}_0}{\rho_0}}=\sqrt{\frac{3\times1.01\times10^5}{1.424}}\text{ms}^{-1}$

$=4.61\times10^2\text{ms}^{-1}$

Now c_rms is also given by

$\text{c}_{\text{rms}}=\sqrt{\frac{3\text{kT}}{\text{m}}}$

$\therefore\frac{\text{c}_{\text{rms}}}{\text{c}_0}=\sqrt{\frac{\text{T}}{\text{T}_0}}$

Here $\text{T}_0=273\text{K}$ and $\text{T}=1092\text{K}$

$\text{c}_{\text{rms}}=\text{c}_0\sqrt{\frac{\text{T}}{\text{T}_0}}$

$=4.61\times10^2\times\sqrt{\frac{1092}{273}}$

$=9.22\times10^2\text{ms}^{-1}$

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Question 473 Marks

A vessel containing one mole of a monatomic ideal gas (molecular weight = 20gmol^-¹) is moving on a floor at a speed of 50ms^-1. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

Answer

N = 1mole, M = 20g/mol, V = 50m/s

K.E. of the vessel = Internal energy of the gas

$=\Big(\frac{1}{2}\Big)\text{mv}^2$

$=\Big(\frac{1}{2}\Big)\times20\times10^{-3}\times50\times50=25\text{J}$

$\Rightarrow25=\text{n}\frac{3}{2}\text{r}(\triangle\text{T})$

$\Rightarrow25=1\times\frac{3}{2}\times8.31\times\triangle\text{T}$

$\Rightarrow\triangle\text{T}=\frac{50}{3\times8.3}\approx2\text{k}$

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Question 483 Marks

Calculate the diameter of a molecule if n = 2.79 × 10²⁵ molecules per m³ and mean free path= 2.2 × 10^-8.

Answer

Here $\text{n}=2.79\times10^{25}\text{molecules m}^{-3};$

$\lambda=2.2\times10^{-8}\text{m}$

Using the relation,

$\lambda=\frac{1}{\sqrt{2}\pi\text{nd}^2}$

We get $\text{d}^2=\frac{1}{\sqrt{2}\pi\text{n}\lambda}$

$=\frac{1}{\sqrt{2}\times3.14\times2.79\times10^{25}\times2.2\times10^{-8}}$

$=0.03666\times10^{-17}\text{m}^2$

$=0.367\times10^{-18}\text{m}^2$

$\Rightarrow\text{d}=\sqrt{0.367\times10^{-18}}\text{m}^2$

$=0.606\times10^{-9}\text{m}=0.606\text{nm}$

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Question 493 Marks

Explain the concept of absolute zero of temperature on the basis of kinetic theory of gases.

Answer

Let us consider one gram mole of the gas. Let M and V be its mass and volume. The pressure exerted by the gas is given by

$\text{P}=\frac{1}3{}\text{mnv}^{-2}$ [M = mn]

$\text{PV}=\frac{1}{3}\text{n}\text{Vmv}^{-2}$

$\text{PV}=\frac{2}{3}\text{N}\times\frac{1}{2}\text{mv}^{-2}$

But $\text{PV = RT}$

$\text{RT}=\frac{2}{3}\text{N}\times\frac{1}{2}\text{mv}^{-2}$

$\frac{1}{2}\text{mv}^{-2}=\frac{3}{2}\frac{\text{R}}{\text{N}}\text{T}$ $\Big[\frac{\text{R}}{\text{N}}=\text{K}_{\text{B}'}\text{Boltzmann Constant}\Big]$

$\frac{1}{2}\text{mv}^{-2}=\frac{3}{2}\text{K}_{\text{B}}\text{T}$

$\bar{\text{v}}\propto\sqrt{\text{T}}$

Hence, absolute temperature is the temperature at which the root mean square velocity of gas molecules reduces to zero.

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Question 503 Marks

Explain why.
There is fall in temperature with increase in altitude.

Answer

The temperature of atmosphere is due to the kinetic energy of air molecule.
Due to lower atmospheric pressure at higher altitude molecules of air rises up so their potential energy increase in turn the kinetic energy decrease results the decrease in temperature.
Due to lower atmospheric pressure at higher altitude the gas expands and gives cooling effect and so decrease the temperature.

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Question 513 Marks

A vessel is filled with a gas at a pressure of 76cm of Hg at a certain temperature. The mass of the gas is increased by 50% by introducing more gas in the vessel at the same temperature. Find out the resultant pressure of the gas.

Answer

According to kinetic theory of gases, the pressure exerted by a gas is

$\text{P}=\frac{1}{3}\rho\text{c}^2=\frac{1}{3}\frac{\text{M}}{\text{V}}\text{c}^2$

As temperature T is kept constant, therefore, c² is constant.

Also, V is constant.

$\therefore\text{P}\propto\text{M}$ or $\frac{{\text{P}_2}}{\text{P}_1}=\frac{\text{M}_2}{\text{M}_1}$

$\frac{\text{P}_2}{76}=\frac{\Big(\text{M}_1+\frac{50}{100}\text{M}_1\Big)}{\text{M}_1}=\frac{3}{2}$

$\text{P}_2=\frac{3}{2}\times76=144\text{cm}$ of mercury (Hg).

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Question 523 Marks

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m³ at a temperature of 27°C and 1 atm pressure.

Answer

Volume of the room, V = 25.0m³

Temperature of the room, T = 27°C = 300K

Pressure in the room, P = 1 atm = 1 × 1.013 × 10⁵Pa

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

PV = k_BNT

Where,

K_B is Boltzmann constant = 1.38 × 10^-23m²kg s^-2k^-1

N is the number of air molecules in the room

$\therefore\text{ N}=\frac{\text{PV}}{\text{K}_{\text{B}}\text{T}}$

$=\frac{1.013\times10^{5}\times25}{(1.38\times10^{23}\times300)}$

= 6.11 × 10²⁶ molecules

Therefore, the total number of air molecules in the given room is 6.11 × 10²⁶

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Question 533 Marks

Calculate (i) r.m.s. velocity and (ii) mean kinetic energy of one gram molecule of hydrogen at S.T.P. Given density of hydrogen at S.T.P. is 0.09kg-m^-3.

Answer

Here, $\rho=0.09\text{kg-m}^{-3}$

At S.T.P., pressure $\text{P}=1.01\times10^5\text{Pa}$

According to kinetic theory of gases.

$\text{P}=\frac{1}{3}\rho\text{C}^2$

$\text{C}=\sqrt{\frac{3\text{P}}{\rho}}$

$=\sqrt{\frac{3\times1.01\times10^5}{0.09}}=1837.5\text{ms}^{-1}$

Volume occupied by one mole of hydrogen at S.T.P. = 22.4 liters = 22.4 × 10^-3 m³

$\therefore$ Mass of hydrogen, M = volume x density

$= 22.4 × 10^{-3} × 0.09$

$= 2.016 × 10^{-3}\text{kg}$

Average K.E./ mole $=\frac{1}2{}\text{MC}^2$

$=\frac{1}{2}\times(2.016\times10^{-3})\\\times(1837.5)^2=3403.4\text{J}$

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Question 543 Marks

Three moles of a diatomic gas is mixed with two moles of monoatomic gas. What will be the molecular specific heat of the mixture at constant volume? [given, R = 8.31J-mol^-1K^-1]

Answer

For a monoatomic gas, i e. $\gamma=\frac{5}{3}$

$\text{C}_{\text{V}_{\gamma}}=\frac{\text{R}}{\gamma-1}=\frac{\text{R}}{\frac{5}{3}-1}=\frac{3}{2}\text{R}$

For a diatomic gas, i.e. $\gamma=\frac{7}{5}$

$\text{C}_{\text{V}}=\frac{\text{R}}{\frac{7}{5}-1}=\frac{5}{2}\text{R}$

By conservation of energy,

$\text{C}_{\text{V}_{\text{mixture}}}=\frac{\mu_1\text{C}_{\text{V}_1}+\mu_2\text{C}_{\text{V}_2}}{\mu_1+\mu_2}$

$=\frac{2\times\frac{3}{2}\text{R}+3\times\frac{5}{2}\text{R}}{2+3}$

$=\frac{3\text{R}+7.5\text{R}}{5}=2.1\text{R}$

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Question 553 Marks

Find the kinetic energy of 1g of nitrogen gas at 77°C. Given, R = 8.31J-mol^-1K^-1.

Answer

For, nitrogen, M = 28

T = 77 + 273 = 350K

R = 8.31J-mol^-1K^-1

K.E. of 1g of nitrogen

$=\frac{3}{2}\frac{\text{RT}}{\text{M}}=\frac{3\times8.31\times350}{2\times50}=155.8\text{J}$

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Question 563 Marks

Molar volume is the volume occupied by 1mol of any (ideal) gas at standard temperature and pressure:
(STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.

Answer

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

PV = nRT

Where,

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

n = Number of moles = 1

T = Standard temperature = 273K

P = Standard pressure = 1atm = 1.013 × 10⁵Nm^-2

$\therefore\ \text {V}=\frac{\text{nRT}}{\text{P}}$

$=\frac{1\times8.314\times273}{(1.013\times10^{5})}$

= 0.0224m³

= 22.4 litres

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

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Question 573 Marks

Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Answer

Pressure, $\text{P}=\frac{1}{3}\frac{\text{nmc}^2}{\text{V}}$

Multiplying and dividing the right side by 2,

$\text{PV}=\frac{2}{3}\frac{1}{2}\text{nmc}^2=\frac{2}{3}\text{E}$

where E is the average kinetic energy.

Also, PV = RT for 1 mole of a gas

$\therefore\text{RT}=\frac{2}{3}\text{E}$

$\therefore$ Kinetic energy (average) is directly proportional to the absolute temperature.

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Question 583 Marks

Consider a sample of oxygen at 300K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth.

Answer

$\text{V}_\text{avg}=\sqrt{\frac{8\text{RT}}{\pi\text{M}}}=\sqrt{\frac{8\times8.3\times300}{3.14\times0.032}}=445.25\text{m/s}$

$\text{T}=\frac{\text{Distance}}{\text{Speed}}=\frac{6400000\times2}{445.25}=445.25\text{m/s}$

$=\frac{28747.83}{3600}\text{km}=7.985\approx8\text{hrs}.$

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Question 593 Marks

The density of an ideal gas is 1.25 × 10^-3gcm³at STP. Calculate the molecular weight of the gas.

Answer

$\text{P}=\frac{\text{nRT}}{\text{V}}=\frac{\text{m}}{\text{M}}\times\frac{\text{RT}}{\text{V}}=\frac{\text{fRT}}{\text{M}}$

ƒ → 1.25 × 10^-3g/ cm³

R → 8.31 × 10⁷ert/ deg/ mole

T → 273K

$\Rightarrow\text{M}=\frac{\text{fRT}}{\text{P}}=\frac{1.25\times10^{-3}\times8.31\times10^7\times273}{13.6\times980\times76}$

$=0.002796\times10^4\approx28\text{g/ mol}$

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Question 603 Marks

The temperature and the relative humidity are 300K and 20% in a room of volume 50m³. The floor is washed with water, 500g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300K = 3.3kPa.

Answer

$\text{RH}=\frac{\text{VP}}{\text{SVP}}\Rightarrow0.20=\frac{\text{VP}}{303\times10^3}$

$\Rightarrow\text{VP}=0.2\times3.3\times10^3=660$

$\text{PV}=\text{nRT}\Rightarrow\text{P}=\frac{\text{nRT}}{\text{V}}=\frac{\text{m}}{\text{M}}\times\frac{\text{RT}}{\text{V}}$

$=\frac{500}{18}\times\frac{8.3\times300}{50}=1383.3$

Net, $\text{P}=1383.3+660=2043.3$

Now, $\text{RH}=\frac{2034.3}{3300}=0.619\approx62\%$

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Question 613 Marks

Why the molecular motion of the molecules ceases at zero kelvin?

Answer

We know, kinetic energy of a molecule is proportional to the absolute temperature.

i.e., $\frac{1}{2}\text{mv}^2_\text{rms}\propto\text{T}$

At $\text{T}=0$

$\frac{1}{2}\text{mv}^2_{\text{rms}}=0,$ Since $\frac{1}{2}\text{m}\neq0,\therefore\text{v}_{\text{rms}}=0$

Thus molecular motion ceases at zero kelvin.

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Question 623 Marks

Isothermal curves for a given mass of gas are shown at two different temperatures T₁ and T₂. State whether T₁ > T₂ or T₂ > T₁. Justify your answer.

Answer

From ideal gas equation

$\text{PV}=\mu\text{RT}$

$\text{T}=\frac{\text{PV}}{\mu\text{R}}$

As mass of gas is constant, μ is constant, R is already a constant.

$\therefore\text{T}\propto\text{PV}$

Since PV is greater for the curve at T₂ than for the curve at T₁, therefore, T₂ > T₁.

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Question 633 Marks

The condition of air in a cloeed room is described as follows. Temperature = 25°C, relative humidity = 60%, pressure = 104kPa. lf all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25°C = 3.2kPa.

Answer

T = 25°C P = 104KPa

$\text{RH}=\frac{\text{VP}}{\text{SVP}}$[SVP = 3.2KPa, RH = 0.6]

$\text{VP} = 0.6 \times 3.2 \times 10^3 = 1.92 \times 10^3 \approx 2 × 10^3$

When vapours are removed VP reduces to zero

Net pressure inside the room now = 104 × 10³ - 2 × 10³ = 102 × 10³ = 102KPa.

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Question 643 Marks

A vertical cylinder of height 100cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.

Answer

P₁ = Atmospheric pressure = 75 × fg

V₁ = 100 × A

P₂ = Atmospheric pressure + Mercury pessue = 75fg + hgfg (if h = height of mercury)

V₂ = (100 - h)A

P₁V₁ = P₂V₂

⇒ 75fg(100A) = (75 + h) fg(100 - h)A

⇒ 75 × 100 = (74 + h)(100 - h)

⇒ 7500 = 7500 - 75h + 100h - h²

⇒ h² - 25h = 0

⇒ h² = 25h

⇒ h = 25cm

Height of mercury that can be poured = 25cm.

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Question 653 Marks

The velocities of ten particles in ms^-1 are 0, 2, 3, 4, 4, 4, 5, 5, 6, 9. Calculate (i) Average speed and (ii) r.m.s. speed.

Answer

Average speed:

$\text{v}_{\text{av}}=\frac{0+2+3+4+4+4+5+5+6+9}{10}=\frac{42}{10}=4.2$

R.M.S. speed:

$\text{v}_{\text{rms}}=\Big[\frac{(0)^2+(2)^2+(3)^2+(4)^2+(4)^2+(4)^2+(5)^2+(5)^2+(6)^2+(9)^2}{10}\Big]^{\frac{1}{2}}$

$=\Big[\frac{228}{10}\Big]^{\frac{1}{2}}=4.77\text{ms}^{-1}$

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Question 663 Marks

Write the difference between ideal gas and real gas.

Answer

	Ideal Gas	Real Gas
(i)	It obeys ideal gas equation, $\text{pV}=\mu\text{RT}$ at all temperatures and pressures	It does not obey, $\text{pV}=\mu\text{RT}$ at all values of temperature and pressure.
(ii)	The volume of the molecules of an ideal gas is zero.	The volume of the molecules of a real gas is non-zero.
(iii)	There is no intermolecular force between the molecules.	There is intermolecular force of attraction or repulsion depending on whether intermolecular separation is larger or small.
(iv)	There is no intermolecular potential energy (U) because intermolecular force (F) is zero.	Potential energy (U) does not equal to zero as intermolecular force (F) is not zero.
(v)	It has only kinetic energy.	It has both kinetic and potential energy.
(vi)	At absolute zero, the volume, pressure and internal energy become zero.	All real gases get liquified before reaching absolute zero. The internal energy of the liquified gas is not zero.

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Question 673 Marks

A vessel contains two non-reactive gases: neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3 : 2. Estimate the ratio of (i) number of molecules, (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2u, molecular mass of O₂ = 32.0u.

Answer

It is given that $\frac{\text{P}_{\text{Ne}}}{\text{P}_{\text{O}_2}}=\frac{3}{2}$

As $\text{P}=\frac{\pi\text{RT}}{\text{V}}$

Hence for a given temp€rafure and volume of vessel,

$\frac{\text{P}_{\text{Ne}}}{\text{P}_{\text{O}_2}}=\frac{\mu_{\text{Ne}}}{\mu_{\text{O}_2}}=\frac{\text{N}_{\text{Ne}}}{\text{N}_{\text{O}_2}}=\frac{3}{2}$

If $\text{m}_{\text{Ne}}$ and $\text{m}_{\text{O}_2}$ be the actual masses of the two gases present in the mixture and $\text{M}_{\text{Ne}}$ and $\text{M}_{\text{O}_2}$ be their molecular masses, then $\mu_{\text{Ne}}=\frac{\text{m}_{\text{Ne}}}{\text{M}_{\text{Ne}}}$ and $\mu_{\text{O}_2}=\frac{\text{m}_{\text{O}_2}}{\text{M}_{\text{O}_2}}.$

If $\rho_{\text{Ne}}$ and $\rho_{\text{O}_2}$ be the mass densities of two gases, then

$\frac{\rho_{\text{Ne}}}{\rho_{\text{O}_2}}=\frac{\text{m}_{\text{Ne}}/{\text{V}}}{\text{m}_{\text{O}_2}/{\text{V}}}=\frac{\text{m}_{\text{Ne}}}{\text{m}_{\text{O}_2}}\times\frac{\text{M}_{\text{Ne}}}{\text{M}_{\text{O}_2}}$

$=\frac{3}{2}\times\frac{20.2}{32.0}=\frac{0.947}{1}$

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Question 683 Marks

A vessel contains 1.60g of oxygen and 2.80g of nitrogen. The temperature is maintained at 300K and the volume of the vessel is 0.166m³. Find the pressure of the mixture.

Answer

$\text{P}_{\text{O}_2}=\frac{\text{n}_{\text{O}_2}\text{RT}}{\text{V}},\text{P}_{\text{H}_2}=\frac{\text{n}_{\text{H}_2}\text{RT}}{\text{V}}$

$\text{n}_{\text{O}_2}=\frac{\text{m}}{\text{M}_{\text{O}_2}}=\frac{1.60}{32}=0.05$

Now, $\text{P}_\text{mix}=\Big(\frac{\text{n}_{\text{O}_2}+\text{n}_{\text{H}_2}}{\text{V}}\Big)\text{RT}$

$\text{n}_{\text{H}_2}=\frac{\text{m}}{\text{M}_{\text{H}_2}}=\frac{2.80}{28}=0.1$

$\text{P}_\text{mix}=\frac{(0.05+0.1)\times8.3\times300}{0.166}=2250\text{N/m}^2$

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Question 693 Marks

A uniform tube closed at one end, contains a pellet of mercury 10cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20cm. Find the length of the air column trapped when the tube is inverted so that the closedend goes down. Atmospheric pressure = 75cm of mercury.

Answer

Case I → Net pressure on air in volume V

= P_atm - hfg = 75 × f_Hg - 10f_Hg = 65 × f_Hg × g

Case II → Net pressure on air in volume ‘V’ = P_atm + f_Hg × g × h

P₁V₁ = P₂V₂

⇒ f_Hg × g × 65 × A × 20 = f_Hg × g × 75 + f_Hg × g × 10 × A × h

⇒ 62 × 20 = 85h

$\Rightarrow\text{h}=\frac{65\times20}{85}=15.2\text{cm}\approx15\text{cm}$

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Question 703 Marks

A vessel A contains hydrogen and another vessel B whose volume is twice of A contains same mass of oxygen at the same temperature. Compare (i) average kinetic energies of hydrogen and oxygen molecules (ii) root mean square speeds of the molecules (iii) pressure of gases in A and B. Molecular weights of hydrogen and oxygen are 2 and 32 respectively.

Answer

For all gases at the same temperature, the kinetic energy per molecule is the same. So, the ratio of the average kinetic energies of hydrogen and oxygen molecules is 1 : 1.
$\frac{\text{C}_1}{\text{C}_2}=\sqrt{\frac{\text{M}_2}{\text{M}_1}}=\sqrt{\frac{32}2{}}=4:1$
we know that $\text{P}=\frac{1}3{}\frac{\text{M}}{\text{V}}\text{C}^2$

In the given problem, M is constant

$\therefore\frac{\text{P}_1}{\text{P}_2}=\frac{\text{C}^2_1}{\text{V}_1}\times\frac{\text{V}_2}{\text{C}^2_2}=\frac{\text{V}_2}{\text{V}_1}\Big[\frac{\text{C}_1}{\text{C}_2}\Big]^2$ or $\frac{\text{P}_1}{\text{P}_2}=\frac{2}{1}\times\frac{16}{1}=32$

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