Characteristics and Measurable properties of gases Questions in English

(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r_1 / r_2 = \sqrt{M_2 / M_1}$.
Here,$r_{He} / r_{CH_4} = n$,$M_{He} = 4 \ g/mol$,and $M_{CH_4} = 16 \ g/mol$.
Substituting the values: $n = \sqrt{16 / 4} = \sqrt{4} = 2$.
Therefore,the value of $n$ is $2$.

(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given that the hydrocarbon $(r_h)$ diffuses $3\sqrt{3}$ times slower than hydrogen $(r_{H_2})$,we have $\frac{r_{H_2}}{r_h} = 3\sqrt{3}$.
Substituting the molar masses ($M_{H_2} = 2 \ g/mol$ and $M_h$ for the hydrocarbon): $\frac{r_{H_2}}{r_h} = \sqrt{\frac{M_h}{M_{H_2}}} = 3\sqrt{3}$.
Squaring both sides: $\frac{M_h}{2} = (3\sqrt{3})^2 = 9 \times 3 = 27$.
Therefore,$M_h = 27 \times 2 = 54 \ g/mol$.
Checking the molar masses of the options:
$A: C_2H_4 = 2(12) + 4(1) = 28 \ g/mol$
$B: C_3H_6 = 3(12) + 6(1) = 42 \ g/mol$
$C: C_4H_6 = 4(12) + 6(1) = 48 + 6 = 54 \ g/mol$
$D: C_6H_6 = 6(12) + 6(1) = 78 \ g/mol$.
The hydrocarbon with molar mass $54 \ g/mol$ is $C_4H_6$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t}$,where $V$ is the volume and $t$ is the time.
Also,$r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.
Therefore,$\frac{V_1 / t_1}{V_2 / t_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $V_{H_2} = 40 \ mL$,$t_{H_2} = 25 \ min$,$M_{H_2} = 2 \ g/mol$.
$V_{O_2} = 20 \ mL$,$M_{O_2} = 32 \ g/mol$.
Substituting the values: $\frac{40 / 25}{20 / t_{O_2}} = \sqrt{\frac{32}{2}}$.
$\frac{40}{25} \times \frac{t_{O_2}}{20} = \sqrt{16} = 4$.
$1.6 \times \frac{t_{O_2}}{20} = 4$.
$t_{O_2} = \frac{4 \times 20}{1.6} = \frac{80}{1.6} = 50 \ min$.

(D) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For two gases,the ratio of their rates of diffusion is given by: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Here,$M_1 = 128 \ g/mol$ (gas) and $M_2 = 32 \ g/mol$ (oxygen).
Substituting the values: $\frac{r_{gas}}{r_{O_2}} = \sqrt{\frac{32}{128}} = \sqrt{\frac{1}{4}} = 0.5$.
Thus,the rate of diffusion of the gas is $0.5$ times that of oxygen.

(A) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the density $(d)$:
$r \propto \frac{1}{\sqrt{d}}$
Therefore,$\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}}$
Given that $r_A = 5 \times r_B$,we have $\frac{r_A}{r_B} = 5$.
Substituting this into the equation:
$5 = \sqrt{\frac{d_B}{d_A}}$
Squaring both sides:
$25 = \frac{d_B}{d_A}$
Thus,$\frac{d_A}{d_B} = \frac{1}{25}$
The ratio of the densities of $A$ and $B$ is $1 : 25$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t}$,where $V$ is volume and $t$ is time.
Graham's Law states: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$,where $M$ is the molar mass.
Substituting $r = \frac{V}{t}$,we get: $\frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $V_1 = 50 \ mL$,$t_1 = 20 \ min$,$M_1 (H_2) = 2 \ g/mol$,$V_2 = 40 \ mL$,$M_2 (O_2) = 32 \ g/mol$.
$\frac{50/20}{40/t_2} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
$\frac{50}{20} \times \frac{t_2}{40} = 4$.
$2.5 \times \frac{t_2}{40} = 4$.
$t_2 = \frac{4 \times 40}{2.5} = \frac{160}{2.5} = 64 \ min$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its density $(d)$: $r \propto \frac{1}{\sqrt{d}}$.
Given the ratio of densities $\frac{d_1}{d_2} = \frac{1}{16}$.
The ratio of their rates of diffusion is $\frac{r_1}{r_2} = \sqrt{\frac{d_2}{d_1}}$.
Substituting the values: $\frac{r_1}{r_2} = \sqrt{\frac{16}{1}} = \frac{4}{1}$.
Therefore,the ratio of their rates of diffusion is $4 : 1$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its density $(d)$ at constant pressure.
$r \propto 1/\sqrt{d}$.
However,the rate of diffusion is also directly proportional to the pressure $(p)$ of the gas.
Combining these,$r \propto p/\sqrt{d}$.
Therefore,the rate of diffusion is directly proportional to $p/\sqrt{d}$.

(B) According to Graham's Law of Effusion,the rate of effusion $r$ is directly proportional to pressure $P$ and inversely proportional to the square root of molar mass $M$: $r \propto \frac{P}{\sqrt{M}}$.
Since $r = \frac{n}{t}$,we have $\frac{n_1}{t_1} \propto \frac{P_1}{\sqrt{M_1}}$ and $\frac{n_2}{t_2} \propto \frac{P_2}{\sqrt{M_2}}$.
Given for $N_2$: $P_1 = 0.8 \, atm$,$t_1 = 38 \, s$,$M_1 = 28 \, g/mol$,$n_1 = 1 \, mol$ (assuming equal moles for comparison).
Given for gas $X$: $P_2 = 1.6 \, atm$,$t_2 = 57 \, s$,$M_2 = ?$,$n_2 = 1 \, mol$.
Using the ratio: $\frac{r_1}{r_2} = \frac{P_1}{P_2} \times \sqrt{\frac{M_2}{M_1}} \times \frac{t_2}{t_1}$.
$\frac{1/38}{1/57} = \frac{0.8}{1.6} \times \sqrt{\frac{M_2}{28}} \times \frac{57}{38}$.
$\frac{57}{38} = 0.5 \times \sqrt{\frac{M_2}{28}} \times \frac{57}{38}$.
$1 = 0.5 \times \sqrt{\frac{M_2}{28}}$.
$2 = \sqrt{\frac{M_2}{28}}$.
$4 = \frac{M_2}{28}$.
$M_2 = 4 \times 28 = 112 \, g/mol$.
(Note: Re-evaluating the calculation: $1.5 = 0.5 \times \sqrt{M_2/28} \times 1.5 \implies 1 = 0.5 \times \sqrt{M_2/28} \implies 2 = \sqrt{M_2/28} \implies 4 = M_2/28 \implies M_2 = 112$. Since $112$ is not in options,checking the ratio again: $r_1/r_2 = (P_1/P_2) \sqrt{M_2/M_1} = (t_2/t_1)$. $(0.8/1.6) \sqrt{M_2/28} = 57/38 = 1.5$. $0.5 \sqrt{M_2/28} = 1.5 \implies \sqrt{M_2/28} = 3 \implies M_2/28 = 9 \implies M_2 = 252 \, g/mol$.)

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Given that the rate of diffusion of $H_2$ is $4$ times that of the mixture,we have $\frac{r_{H_2}}{r_{mix}} = 4$.
Therefore,$\frac{\sqrt{M_{mix}}}{\sqrt{M_{H_2}}} = 4$,which implies $\frac{M_{mix}}{M_{H_2}} = 16$.
Since $M_{H_2} = 2 \ g/mol$,the average molar mass of the mixture $M_{mix} = 16 \times 2 = 32 \ g/mol$.
Let the mole fraction of $C_2H_4$ be $x$ and that of $CO_2$ be $(1-x)$.
The molar mass of $C_2H_4$ is $28 \ g/mol$ and $CO_2$ is $44 \ g/mol$.
$28x + 44(1-x) = 32$.
$28x + 44 - 44x = 32$.
$-16x = -12$,so $x = \frac{12}{16} = 0.75$.
The mole fraction of $C_2H_4$ is $0.75$ and $CO_2$ is $0.25$.
The molar ratio $C_2H_4 : CO_2 = 0.75 : 0.25 = 3 : 1$.

(D) Graham's Law of diffusion states that the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$ or density $(d)$: $r \propto \frac{1}{\sqrt{M}}$.
The rate of diffusion $(r)$ can be expressed as:
$r = \frac{V}{t}$ (Volume diffused per unit time)
$r = \frac{n}{t}$ (Number of moles/molecules diffused per unit time)
$r = \frac{m}{t}$ (Mass diffused per unit time)
However,the decrease in pressure per unit time is not a standard definition used in the calculation of the rate of diffusion in Graham's Law.

(D) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} = \frac{k}{\sqrt{M}}$,where $V$ is volume,$t$ is time,and $M$ is molar mass.
Since the time $t$ is the same for all gases,the volume $V$ diffused is directly proportional to $\frac{1}{\sqrt{M}}$.
Therefore,$\frac{V_1}{\sqrt{M_1}} = \frac{V_2}{\sqrt{M_2}}$.
For $H_2$: $V_1 = 64 \ mL$,$M_1 = 2 \ g/mol$.
$\frac{64}{\sqrt{2}} = \frac{V_2}{\sqrt{M_2}} \implies V_2 = \frac{64 \times \sqrt{M_2}}{\sqrt{2}} = 64 \times \sqrt{\frac{M_2}{2}}$.
Checking option $(D)$ for $He$ $(M_2 = 4)$:
$V_2 = 64 \times \sqrt{\frac{4}{2}} = 64 \times \sqrt{2} \approx 64 \times 1.414 = 90.5 \ mL$ (Incorrect).
Checking option $(C)$ for $D_2$ $(M_2 = 4)$:
$V_2 = 64 \times \sqrt{\frac{4}{2}} = 90.5 \ mL$ (Incorrect).
Checking option $(B)$ for $CH_4$ $(M_2 = 16)$:
$V_2 = 64 \times \sqrt{\frac{16}{2}} = 64 \times \sqrt{8} = 64 \times 2.828 = 181 \ mL$ (Incorrect).
Re-evaluating the calculation: The rate $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$. For same $t$,$V \propto \frac{1}{\sqrt{M}}$.
$V_{gas} = V_{H_2} \times \sqrt{\frac{M_{H_2}}{M_{gas}}} = 64 \times \sqrt{\frac{2}{M_{gas}}}$.
For $He$ $(M=4)$: $V = 64 \times \sqrt{0.5} = 64 \times 0.707 = 45.26 \ mL$. Thus,option $(D)$ is correct.

(C) At high altitude,the atmospheric pressure is lower than at sea level.
Because the boiling point of a liquid is dependent on the external pressure,liquids boil at lower temperatures at high altitudes.
Since the food is cooked at a lower temperature,it takes more time to cook.
Therefore,a pressure cooker is used to increase the internal pressure,which raises the boiling point of water and facilitates faster cooking.

(B) According to Graham's law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Thus,$\frac{r_g}{r_H} = \sqrt{\frac{M_H}{M_g}}$.
Given that the rate of the gas is $1/3$ times that of hydrogen $(H_2)$,where $M_H = 2 \ g/mol$,we have $\frac{1}{3} = \sqrt{\frac{2}{M_g}}$.
Squaring both sides: $\frac{1}{9} = \frac{2}{M_g}$.
Therefore,$M_g = 18 \ g/mol$.

(C) The viscosity of a liquid is due to the intermolecular forces of attraction between its molecules.
As the temperature increases,the kinetic energy of the molecules increases,which helps them overcome these intermolecular forces.
Therefore,the viscosity of the liquid decreases with an increase in temperature.
$T \uparrow \implies \text{viscosity} \downarrow$

(B) The spherical shape of liquid droplets,such as mercury or rain,is caused by the characteristic property of liquids known as $surface \ tension$.
This property acts to minimize the surface area of the liquid for a given volume,and for a given volume,a sphere has the minimum surface area.

(A) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$ and directly proportional to the volume $V$ effused per unit time $t$:
$r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$
Since the volumes are the same,we have $\frac{V}{t_A} \times \sqrt{M_A} = \frac{V}{t_B} \times \sqrt{M_B}$,which simplifies to $\frac{t_B}{t_A} = \sqrt{\frac{M_B}{M_A}}$.
Given $t_A = 30 \text{ min}$,$t_B = 10 \text{ min}$,and $M_A = 162$.
Substituting the values: $\frac{10}{30} = \sqrt{\frac{M_B}{162}}$.
$\frac{1}{3} = \sqrt{\frac{M_B}{162}}$.
Squaring both sides: $\frac{1}{9} = \frac{M_B}{162}$.
$M_B = \frac{162}{9} = 18$.

(C) During the process of evaporation,the particles of the liquid absorb energy from the surroundings to overcome the forces of attraction and change into the vapor state. This energy is known as the latent heat of vaporization. Since this energy is used for the phase change rather than increasing the kinetic energy of the particles,the temperature of the liquid remains constant during the process.

(A) For a diatomic gas,the degrees of freedom $(f)$ is $5$.
The molar heat capacity at constant volume is $C_v = \frac{f}{2}R = \frac{5}{2}R$.
The molar heat capacity at constant pressure is $C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$.
The ratio of specific heats is $\gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$.

(A) For a monoatomic gas like $He$,the degrees of freedom $f = 3$.
The molar heat capacity at constant volume is $C_v = (f/2)R = (3/2)R$.
Since $C_p = C_v + R$,we have $C_p = (3/2)R + R = (5/2)R$.
Thus,statement $A$ is correct.

(D) According to Graham's law of diffusion,the rate of diffusion $(r)$ of a gas is directly proportional to its pressure $(P)$ and inversely proportional to the square root of its molar mass $(M)$.
Mathematically,this is expressed as $r \propto \frac{P}{\sqrt{M}}$.

(B) According to Graham's Law of effusion,the rate of effusion $r \propto \frac{1}{\sqrt{M}}$.
Since $r = \frac{V}{t}$,for the same volume $V$,we have $\frac{V}{t_1} \times \sqrt{M_1} = \frac{V}{t_2} \times \sqrt{M_2}$,which simplifies to $\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$.
Given $t_1 = 5 \ s$ for $H_2$ $(M_1 = 2 \ g/mol)$,we calculate $t_2$ for each gas:
For $O_2$ $(M_2 = 32 \ g/mol)$: $t_2 = 5 \times \sqrt{\frac{32}{2}} = 5 \times \sqrt{16} = 5 \times 4 = 20 \ s$.
Thus,the correct option is $20 \ seconds : O_2$.

(B) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
First,calculate the molar masses $(M)$ of the given gases:
$M(SO_2) = 32 + 2 \times 16 = 64 \ g/mol$
$M(CO_2) = 12 + 2 \times 16 = 44 \ g/mol$
$M(PCl_3) = 31 + 3 \times 35.5 = 137.5 \ g/mol$
$M(SO_3) = 32 + 3 \times 16 = 80 \ g/mol$
Comparing the molar masses: $M(PCl_3) > M(SO_3) > M(SO_2) > M(CO_2)$.
Since the rate of diffusion is inversely proportional to the square root of the molar mass,the order of the rate of diffusion is: $CO_2 > SO_2 > SO_3 > PCl_3$.

(D) According to the ideal gas equation,$PV = nRT$.
Since the gases are in the same cylinder at the same temperature and volume,$n = \frac{PV}{RT}$.
Therefore,$n \propto P$.
Thus,the ratio of the number of moles is equal to the ratio of their partial pressures:
$\frac{n_{C_3H_6}}{n_{O_2}} = \frac{P_{C_3H_6}}{P_{O_2}} = \frac{170 \ torr}{570 \ torr} = 0.30$.

(C) The total pressure of the saturated air is the sum of the partial pressure of dry air and the partial pressure of water vapour.
Given,the mole fraction of water vapour,$X_{H_{2}O} = 0.02$.
The mole fraction of dry air,$X_{dry\;air} = 1 - X_{H_{2}O} = 1 - 0.02 = 0.98$.
The partial pressure of a component is given by $P_{i} = X_{i} \times P_{total}$.
Therefore,the partial pressure of dry air $= X_{dry\;air} \times P_{total} = 0.98 \times 1.2 \; atm = 1.176 \; atm$.

(N/A) Number of moles of dioxygen $(n_{O_2})$ $= \frac{70.6 \ g}{32 \ g \ mol^{-1}} = 2.21 \ mol$.
Number of moles of neon $(n_{Ne})$ $= \frac{167.5 \ g}{20 \ g \ mol^{-1}} = 8.375 \ mol$.
Total moles $(n_{total})$ $= 2.21 + 8.375 = 10.585 \ mol$.
Mole fraction of dioxygen $(x_{O_2})$ $= \frac{2.21}{10.585} = 0.21$.
Mole fraction of neon $(x_{Ne})$ $= \frac{8.375}{10.585} = 0.79$.
Partial pressure of a gas $= \text{mole fraction} \times \text{total pressure}$.
Partial pressure of dioxygen $= 0.21 \times 25 \ bar = 5.25 \ bar$.
Partial pressure of neon $= 0.79 \times 25 \ bar = 19.75 \ bar$.

(N/A) The physical state of matter is determined by the competition between two opposing forces: intermolecular forces and thermal energy.
$1$. Intermolecular forces: These forces tend to keep the molecules together,promoting the formation of condensed phases like liquids and solids.
$2$. Thermal energy: This is the energy of molecular motion,which arises due to the temperature of the substance. It tends to keep the molecules apart,promoting the gaseous state.
The three states of matter are the result of the balance between these two factors:
- When intermolecular forces are dominant,the substance exists as a solid or liquid.
- When thermal energy is dominant,the substance exists as a gas.
As shown in the diagram,the transition from gas to solid is driven by the increasing dominance of intermolecular interactions,while the transition from solid to gas is driven by the increasing dominance of thermal energy. Gases cannot be liquefied by compression alone if the thermal energy is high; lowering the temperature reduces the thermal energy,allowing intermolecular forces to hold the molecules together,thereby facilitating liquefaction.

(D) The gaseous state is characterized by the following properties:
$1$. Gases are highly compressible because of large intermolecular spaces.
$2$. Gases exert pressure equally in all directions due to the continuous collision of molecules with the walls of the container.
$3$. Gases have much lower density than solids and liquids because the molecules are far apart.
$4$. The volume and the shape of gases are not fixed; they assume the volume and shape of the container.
$5$. Gases mix evenly and completely in all proportions without any mechanical aid.

(N/A) The simplicity of the behavior of gases is due to the fact that the forces of interaction between their molecules are negligible.
Their behavior is governed by the same common general laws,which were discovered as a result of their experimental studies.
These laws represent relationships between measurable properties of gases. Some of these properties,such as pressure $(P)$,volume $(V)$,temperature $(T)$,and mass $(m)$,are very important because the relationships between these variables describe the state of the gas.

(B) barometer is an instrument used to measure atmospheric pressure.
$A$ manometer is an instrument used to measure the pressure of a gas in a closed container.
Example: If the atmospheric pressure (barometer reading) is $740 \ mm \ Hg$ and the difference in height between the two ends of a manometer is $2.1 \ cm$,the pressure of the gas can be calculated as:
Pressure of gas = Atmospheric pressure - Difference in height
$= 740 \ mm \ Hg - (2.1 \ cm \times 10 \ mm/cm)$
$= 740 \ mm \ Hg - 21 \ mm \ Hg$
$= 719 \ mm \ Hg$

(N/A) Matter exists in three states: solid,liquid,and gas.
The stability of a particular state of matter at a given temperature and pressure depends on the net effect of two opposing factors: intermolecular forces and thermal energy.
Intermolecular forces tend to keep the molecules (or atoms or ions) closer together,whereas thermal energy tends to keep them apart by making them move faster.
When intermolecular forces are strong enough to hold the particles together at fixed positions,the substance exists in the solid state.
In this state,particles only oscillate about their mean positions.

(N/A) The stability of the state of matter depends on two opposing factors:
$(i)$ Intermolecular forces
$(ii)$ Thermal energy
Intermolecular forces tend to keep the molecules,atoms,or ions closer,while thermal energy tends to keep them apart by making them move faster.
At a sufficiently low temperature,the thermal energy is low,and intermolecular forces bring the particles so close that they cling to one another and occupy fixed positions. These particles can still oscillate about their mean positions,and the substance exists in the solid state.

(N/A) The time taken for complete evaporation of a liquid depends on the following three factors:
$1$. The nature of the liquid: Different liquids have different rates of evaporation at a constant temperature. For example,liquids like ether,acetone,and petrol evaporate faster than water.
$2$. Amount of liquid: At a constant temperature,the time required for complete evaporation is directly proportional to the amount of liquid. For instance,$10 \ mL$ of water takes more time to evaporate completely than $5 \ mL$ of water.
$3$. Temperature: The rate of evaporation increases with an increase in temperature. Therefore,at a constant pressure of $1 \ \text{atm}$,a given amount of liquid will evaporate in less time at a higher temperature.

(N/A) Boiling point: The temperature at which the vapour pressure of a liquid becomes equal to the external atmospheric pressure (typically $1 \text{ atm}$) is called the boiling point of the liquid.
Factors affecting boiling point:
$1$. Nature of liquid: Each liquid has a characteristic boiling point. If the intermolecular forces of attraction in a liquid are weak,the molecules can easily escape into the vapour phase,resulting in higher vapour pressure and a lower boiling point. Such liquids are more volatile. For example,the boiling points of water,ethanol,and ether are $100^{\circ}C$,$78^{\circ}C$,and $34^{\circ}C$ respectively.
$2$. External pressure: The boiling point of a liquid depends on the external pressure applied to its surface. If the atmospheric pressure is lower,the liquid reaches its boiling point at a lower temperature. For example,on high mountains,atmospheric pressure is lower,so water boils at a temperature lower than $100^{\circ}C$. Conversely,in a pressure cooker,the pressure is increased,which raises the boiling point of water,allowing food to cook faster.

(N/A) The physical states of matter can be changed by altering factors such as $Temperature$ and $Pressure$.

(N/A) Absolute zero temperature: The temperature $-273.15^{\circ}C$ (approximately $-273^{\circ}C$) is defined as absolute zero temperature.
It is denoted by $T$ and its unit is $K$ (Kelvin). The degree symbol $(^{\circ})$ is not used with Kelvin.
Absolute temperature in Kelvin is calculated as: $T(K) = 273.15 + t(^{\circ}C)$.
For simplicity,it is often written as: $T(K) = 273 + t(^{\circ}C)$.
Definition: The scale of temperature starting from absolute zero is called the Kelvin scale of temperature or the thermodynamic scale.

(N/A) $(i)$ Isotherm: $A$ graph plotted between pressure and volume at a constant temperature is known as an isotherm.
Example: Boyle's Law.
$(ii)$ Isochore: $A$ graph plotted between pressure and temperature at a constant volume is known as an isochore.
Example: Gay-Lussac's Law.
$(iii)$ Isobar: $A$ graph plotted between volume and temperature at a constant pressure is known as an isobar.
Example: Charles's Law.

(N/A) Dalton's Law: Formulated by John Dalton in $1801$,it states that the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of the individual gases.
Partial pressure $(p)$: In a mixture of gases,the pressure exerted by an individual gas is called its 'partial pressure'.
Mathematical formula: If $p_{\text{Total}}$ is the total pressure and $p_{1}, p_{2}, p_{3}, \dots$ are the partial pressures of individual gases at constant $T$ and $V$:
$p_{\text{Total}} = p_{1} + p_{2} + p_{3} + \dots$ (Eq.-$I$)
Pressure of dry gas: Gases collected over water are moist. To find the pressure of the dry gas:
$p_{\text{Dry gas}} = p_{\text{Total}} - p_{\text{Aqueous tension}}$ (Eq.-$II$)
Aqueous tension: The pressure exerted by saturated water vapour is called 'aqueous tension'.

Temp./$K$	Pressure/bar
$273.15$	$0.0060$
$283.15$	$0.0121$
$288.15$	$0.0168$
$291.15$	$0.0204$
$293.15$	$0.0230$
$295.15$	$0.0260$
$297.15$	$0.0295$
$299.15$	$0.0331$
$301.15$	$0.0372$
$303.15$	$0.0418$

(N/A) Dalton's law: The law was formulated by John Dalton in $1801$. "It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases."
i.e.,the pressures which these gases would exert if they were enclosed separately in the same volume and under the same conditions of temperature.
Partial pressure $(p):$ In a mixture of gases,the pressure exerted by the individual gas is called 'Partial pressure'.
Mathematical formula of Dalton's Law of partial pressure: Where $p_{\text{Total}}$ is the total pressure exerted by the mixture of gases and $p_{1}, p_{2}, p_{3}$ etc. are partial pressures of gases.
$p_{\text{Total}} = p_{1} + p_{2} + p_{3} + \ldots$ (at constant $T$ and $V$) ....(Eq.-$i$)
Pressure of dry gas: Gases are generally collected over water and therefore are moist. So we have to obtain the pressure of dry gas.
$\left(\text{Pressure of Dry gas}\right) = \left(\text{Total pressure of moist gas contains water vapours}\right) - \left(\text{Vapour pressure of water}\right)$
$\therefore p_{\text{Dry gas}} = (p_{\text{Total}} - p_{\text{Aqueous tension}}) \quad \ldots$ (Eq.-$ii$)
Aqueous tension: Pressure exerted by saturated water vapour is called 'aqueous tension'. Aqueous tension of water at different temperatures is given in table.

Temp./$K$	Pressure/bar
$273.15$	$0.0060$
$283.15$	$0.0121$
$288.15$	$0.0168$
$291.15$	$0.0204$
$293.15$	$0.0230$
$295.15$	$0.0260$
$297.15$	$0.0295$
$299.15$	$0.0331$
$301.15$	$0.0372$
$303.15$	$0.0418$

Necessity to follow Dalton's law: $(i)$ Gases of mixture do not react with each other chemically $(ii)$ Temperature and pressure of different gases remains same.

Suppose at a constant temperature $(T)$, three gases are enclosed in a volume $(V)$, exerting partial pressures $p_{1}, p_{2}$, and $p_{3}$ respectively.
Using the ideal gas equation $pV = nRT$, we have:
$(i) \ p_{1} = \frac{n_{1}RT}{V} \quad \dots (Eq.-i)$
$(ii) \ p_{2} = \frac{n_{2}RT}{V} \quad \dots (Eq.-ii)$
$(iii) \ p_{3} = \frac{n_{3}RT}{V} \quad \dots (Eq.-iii)$
where $n_{1}, n_{2}$, and $n_{3}$ are the number of moles of these gases.
According to Dalton's Law, the total pressure $(p_{\text{total}})$ is:
$p_{\text{total}} = p_{1} + p_{2} + p_{3} = (n_{1} + n_{2} + n_{3}) \frac{RT}{V} \quad \dots (Eq.-iv)$
To find the relation, divide the partial pressure of a gas by the total pressure:
$\frac{p_{1}}{p_{\text{total}}} = \frac{n_{1}RT/V}{(n_{1} + n_{2} + n_{3})RT/V} = \frac{n_{1}}{n_{1} + n_{2} + n_{3}}$
Since the mole fraction $(\chi_{1})$ of the first gas is defined as $\chi_{1} = \frac{n_{1}}{n_{\text{total}}}$, where $n_{\text{total}} = n_{1} + n_{2} + n_{3}$, we get:
$\frac{p_{1}}{p_{\text{total}}} = \chi_{1}$
Therefore, the relation is $p_{1} = \chi_{1} p_{\text{total}}$.

(A) The total number of moles $n_{total} = n_{O_2} + n_{Cl_2} + n_{N_2} = 4 + 3 + 3 = 10 \, mole$.
According to Dalton's Law of Partial Pressure,the partial pressure of a gas is given by $p_i = x_i \times P_{total}$,where $x_i$ is the mole fraction of the gas.
Mole fraction of $O_2$ $(x_{O_2})$ = $4 / 10 = 0.4$.
Mole fraction of $Cl_2$ $(x_{Cl_2})$ = $3 / 10 = 0.3$.
Mole fraction of $N_2$ $(x_{N_2})$ = $3 / 10 = 0.3$.
Partial pressure of $O_2$ $(p_{O_2})$ = $0.4 \times 50 \, bar = 20 \, bar$.
Partial pressure of $Cl_2$ $(p_{Cl_2})$ = $0.3 \times 50 \, bar = 15 \, bar$.
Partial pressure of $N_2$ $(p_{N_2})$ = $0.3 \times 50 \, bar = 15 \, bar$.

(A) According to Dalton's law of partial pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure. Since the volume percentage is equal to the mole fraction for ideal gases,we have: $p_{He} = \chi_{He} \times P_{total} = 0.40 \times 25 \ bar = 10 \ bar$ $p_{Ne} = \chi_{Ne} \times P_{total} = 0.40 \times 25 \ bar = 10 \ bar$ $p_{Ar} = \chi_{Ar} \times P_{total} = 0.20 \times 25 \ bar = 5 \ bar$

$1$. Calculate the moles of $O_2$: $n = \frac{0.32 \, g}{32 \, g/mol} = 0.01 \, mol$.
$2$. Calculate the total pressure $(P_{total})$ using the ideal gas equation $PV = nRT$: $P_{total} = \frac{nRT}{V} = \frac{0.01 \, mol \times 8.314 \times 10^{-2} \, L \, bar \, mol^{-1} \, K^{-1} \times 500 \, K}{2 \, L} = 0.20785 \, bar$.
$3$. Convert water vapour pressure to $bar$: $P_{H_2O} = 32 \, mbar = 0.032 \, bar$.
$4$. Calculate the partial pressure of dry $O_2$ using Dalton's Law: $P_{O_2} = P_{total} - P_{H_2O} = 0.20785 \, bar - 0.032 \, bar = 0.17585 \, bar$.

(A) The total ratio is $1 + 2 + 7 = 10$.
Volume percentage of $Cl_2 = (1/10) \times 100 = 10 \%$.
Volume percentage of $H_2 = (2/10) \times 100 = 20 \%$.
Volume percentage of $N_2 = (7/10) \times 100 = 70 \%$.
Partial pressure of $Cl_2 = (10/100) \times 40 \ bar = 4 \ bar$.
Partial pressure of $H_2 = (20/100) \times 40 \ bar = 8 \ bar$.
Partial pressure of $N_2 = (70/100) \times 40 \ bar = 28 \ bar$.

(1.06 BAR) According to Dalton's Law of partial pressures,the total pressure of a gas collected over water is the sum of the pressure of the dry gas and the vapour pressure of water.
$P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapour}}$
Given:
$P_{\text{total}} = 33.26 \ bar$
$P_{\text{dry } O_2} = 32.20 \ bar$
Therefore,
$P_{\text{water vapour}} = P_{\text{total}} - P_{\text{dry } O_2}$
$P_{\text{water vapour}} = 33.26 \ bar - 32.20 \ bar = 1.06 \ bar$

(A) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is given by the product of its mole fraction and the total pressure.
Since volume percentage is equal to mole fraction in a gaseous mixture,the mole fractions are:
$x_{N_2} = 0.70, x_{O_2} = 0.20, x_{CO_2} = 0.01$.
Given total pressure $P_{total} = 1 \, bar$.
Partial pressure $P_i = x_i \times P_{total}$.
$P_{N_2} = 0.70 \times 1 \, bar = 0.70 \, bar$.
$P_{O_2} = 0.20 \times 1 \, bar = 0.20 \, bar$.
$P_{CO_2} = 0.01 \times 1 \, bar = 0.01 \, bar$.

(N/A) Equilibrium or Saturated Vapour Pressure: If an evacuated container is partially filled with a liquid,a portion of the liquid evaporates to fill the remaining volume of the container with vapour.
Initially,the liquid evaporates and the pressure exerted by the vapours on the walls of the container (vapour pressure) increases. After some time,it becomes constant,and an equilibrium is established between the liquid phase and the vapour phase. The vapour pressure at this stage is known as 'equilibrium vapour pressure' or 'saturated vapour pressure'.
Since the process of vaporisation is temperature-dependent,the temperature must be mentioned while reporting the vapour pressure of a liquid.
Boiling and Boiling Point:
- Boiling: When a liquid is heated in an open vessel,it vaporises from the surface. At the temperature where the vapour pressure of the liquid becomes equal to the external pressure,vaporisation occurs throughout the bulk of the liquid,and vapours expand freely into the surroundings. This condition of free vaporisation throughout the liquid is called boiling.
- Boiling Point: The temperature at which the vapour pressure of a liquid is equal to the external pressure is called the boiling point temperature at that pressure.
Normal and Standard Boiling Point: At $1 \text{ atm}$ pressure,the boiling temperature is called the 'normal boiling point',and at $1 \text{ bar}$,the boiling point is called the 'standard boiling point'. Note that the standard boiling point $(1 \text{ bar})$ is less than the normal boiling point $(1 \text{ atm})$.
e.g.,$\text{Standard boiling point of } H_2O = 99.6^{\circ}C \ (372.6 \ K) < \text{Normal boiling point of } H_2O = 100^{\circ}C \ (373 \ K)$,because $1 \text{ bar} < 1 \text{ atm}$.
Effect of Pressure on Boiling Point: If external pressure decreases,the boiling point decreases; if external pressure increases,the boiling point increases.
Examples:
$1$. On mountains,atmospheric pressure is lower,so the boiling point of water is lower,making it difficult to cook food properly.
$2$. In a pressure cooker,pressure is increased,which increases the boiling point of water,allowing for faster and more efficient cooking.
$3$. In hospitals,surgical instruments are sterilized in autoclaves where the boiling point of water is increased by raising the pressure above atmospheric pressure.

(N/A) Effect of Surface tension: Liquids generally assume the shape of their container. However,some phenomena occur due to surface tension:
$-$ Mercury forms spherical beads instead of spreading on a surface.
$-$ Soil particles at the bottom of a river remain separated but stick together when taken out.
$-$ $A$ liquid rises in a thin capillary tube upon contact.
Explanation of surface tension:
$A$ molecule in the bulk of a liquid experiences equal intermolecular forces from all sides,resulting in no net force. However,a molecule on the surface experiences a net attractive force towards the interior of the liquid because there are no molecules above it. This net downward force gives surface molecules more energy than those in the bulk. Consequently,liquids tend to minimize the number of molecules at their surface.
Surface energy: The energy required to increase the surface area of a liquid by one unit is called 'surface energy' or 'surface tension energy'.
Definition of Viscosity: Viscosity is a measure of the resistance to flow,arising from internal friction between layers of fluid as they slip past one another. Strong intermolecular forces hold molecules together and resist the movement of layers.
Laminar flow: When a liquid flows over a fixed surface,the layer in contact with the surface is stationary. The velocity of upper layers increases with distance from the fixed layer. This regular gradation of velocity is called 'Laminar flow'.
Mathematical Representation of Viscosity:
If the velocity of a layer at a distance $dz$ changes by a value $du$,the velocity gradient is $\frac{du}{dz}$.
The force $F$ required to maintain flow is proportional to the area of contact $A$ and the velocity gradient $\frac{du}{dz}$:
$(i)$ $F \propto A$
$(ii)$ $F \propto \frac{du}{dz}$
$\therefore F \propto A \left( \frac{du}{dz} \right)$
$\therefore F = \eta A \left( \frac{du}{dz} \right)$
Where $\eta$ is the coefficient of viscosity.