Characteristics and Measurable properties of gases Questions in English

(A) One $Pascal$ $(Pa)$ is the pressure exerted by a force of one $newton$ $(N)$ on an area of one square meter $(m^2)$.
Atmospheric pressure is defined as the force per unit area exerted by the weight of the atmosphere.
The standard value for one atmospheric pressure in terms of $Pascal$ is $1 \ atm = 101325 \ Pa$.
Since $1 \ kPa = 1000 \ Pa$,we have $1 \ atm = 101.325 \ kPa$.

(C) The correct answer is $(C)$. Gases do not have a definite shape or a definite volume. Their volume is equal to the volume of the container they occupy,as they expand to fill the entire space available.

(B) In the gaseous state,the kinetic energy (thermal energy) of the molecules is much higher than the intermolecular forces of attraction.
This high thermal energy allows the molecules to overcome the attractive forces,move freely,and occupy the entire available volume.
Therefore,the correct relationship is $Thermal \ energy >> Molecular \ attraction$.

(B) barometer is used to measure atmospheric pressure.
$A$ stalagmometer is used to measure the surface tension of liquids.
$A$ manometer is specifically designed to measure the pressure of a gas confined in a vessel.

(D) The mass of gas can be determined by weighing the container filled with gas and again weighing the container after removing the gas. The difference between the two weights gives the mass of the gas.

(A) . At sea level,the air is compressed by the weight of the atmosphere above it,leading to higher pressure. According to Boyle's law,at a constant temperature,the volume of a given mass of gas is inversely proportional to its pressure $(P \propto 1/V)$. As pressure increases,volume decreases,which results in an increase in density $(d = m/V)$. Therefore,air at sea level is denser.

(D) Dalton's law of partial pressures states that for a mixture of non-reacting gases,the total pressure $(P_{total})$ is the sum of the partial pressures of the individual gases $(P_1, P_2, P_3, \dots)$ at constant temperature and volume. Mathematically,$P_{total} = P_1 + P_2 + P_3 + \dots$.

(A) According to Dalton's law of partial pressure,the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases present in the mixture.
Therefore,$P_{\text{total}} = P_A + P_B + P_C$.
The number of moles given ($1$,$2$,and $3$) is irrelevant to the calculation of total pressure when partial pressures are already provided.

(D) Because $NH_3$ and $HCl$ gases react to form solid $NH_4Cl$ $(NH_3(g) + HCl(g) \rightarrow NH_4Cl(s))$.
Dalton's law of partial pressure is only applicable to mixtures of non-reacting gases.

(B) Dalton's law of partial pressures is applicable only to a mixture of non-reacting gases.
In option $(B)$,$NH_3$ reacts with $HCl$ and $HBr$ to form solid $NH_4Cl$ and $NH_4Br$ respectively.
Since these gases react with each other,Dalton's law is not applicable to this mixture.

(D) 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 $(d)$ or its molar mass $(M)$.
Mathematically,$r \propto \frac{1}{\sqrt{d}}$ or $r \propto \frac{1}{\sqrt{M}}$.
Therefore,the rate of diffusion is inversely proportional to the square root of its molecular mass.

(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 $(d)$ or molar mass $(M)$.
Mathematically,$r \propto \frac{1}{\sqrt{d}}$ or $r \propto \sqrt{\frac{1}{d}}$.

(C) According to Graham's law of diffusion,the rate of diffusion $r$ is directly proportional to the pressure $P$ and inversely proportional to the square root of the molar mass $M$ of the gas.
Mathematically,$r \propto \frac{P}{\sqrt{M}}$.
For two gases $A$ and $B$,the ratio of their rates of diffusion is given by:
$\frac{r_A}{r_B} = \frac{P_A}{P_B} \times \sqrt{\frac{M_B}{M_A}} = \frac{P_A}{P_B} \left(\frac{M_B}{M_A}\right)^{1/2}$

(A) According to Graham's Law of diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given $\frac{r_g}{r_{He}} = 1.4$ and $M_{He} = 4$.
Substituting the values: $1.4 = \sqrt{\frac{4}{M_g}}$.
Squaring both sides: $(1.4)^2 = \frac{4}{M_g}$.
$1.96 = \frac{4}{M_g}$.
$M_g = \frac{4}{1.96} \approx 2.04$.
Since $1.4 \approx \sqrt{2}$,the value is $2$.

(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$: $r \propto \frac{1}{\sqrt{M}}$.
Given that the rate of the gas $r_g = \frac{1}{5} r_{H_2}$,we have $\frac{r_{H_2}}{r_g} = 5$.
Using the relation $\frac{r_{H_2}}{r_g} = \sqrt{\frac{M_g}{M_{H_2}}}$,we get $5 = \sqrt{\frac{M_g}{2}}$.
Squaring both sides,$25 = \frac{M_g}{2}$,which gives $M_g = 50$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$: $r_1/r_2 = \sqrt{M_2/M_1}$.
Given that the rate of the gas $r_g = \frac{1}{6} r_{H_2}$,we have $\frac{r_{H_2}}{r_g} = 6$.
Using the formula $M_g = M_{H_2} \times (\frac{r_{H_2}}{r_g})^2$,where $M_{H_2} = 2 \ g/mol$.
$M_g = 2 \times (6)^2 = 2 \times 36 = 72 \ g/mol$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the density $(d)$:
$r_1 / r_2 = \sqrt{d_2 / d_1}$
Given:
$r_H = 1$
$d_H = 0.09 \ g \ L^{-1}$
$d_O = 1.44 \ g \ L^{-1}$
Substituting the values:
$r_H / r_O = \sqrt{d_O / d_H}$
$1 / r_O = \sqrt{1.44 / 0.09} = \sqrt{16} = 4$
$r_O = 1 / 4 = 0.25$

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the density $d$ of the gas: $r \propto \frac{1}{\sqrt{d}}$.
Therefore,$\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}}$.
Given that $r_A = 5r_B$,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,the ratio of the density of $A$ to the density of $B$ is $\frac{d_A}{d_B} = \frac{1}{25} = 0.04$.

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

(D) According to Graham's Law of diffusion,the rate of diffusion $D$ of a gas is inversely proportional to the square root of its density $\rho$.
$\frac{D_A}{D_B} = \sqrt{\frac{\rho_B}{\rho_A}} = (\frac{\rho_B}{\rho_A})^{1/2}$
Therefore,$D_A = D_B (\frac{\rho_B}{\rho_A})^{1/2}$.

(C) $Atmolysis$ is a technique used for the separation of gases from their gaseous mixture.
This process relies on the difference in the rates of diffusion of gases,which is dependent on their densities (Graham's Law of Diffusion).
Therefore,option $(c)$ is correct.

(A) According to Graham's law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
Given $M_{O_2} = 32 \ g/mol$ and $M_{H_2} = 2 \ g/mol$,we have $\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Since rate $r = \frac{\text{mass}}{t}$,for the same time $t$,$\frac{\text{mass}_{H_2}}{\text{mass}_{O_2}} = 4$.
Given $\text{mass}_{O_2} = 4 \ g$,then $\text{mass}_{H_2} = 4 \times 4 = 16 \ g$.

(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Since the time $t$ is the same for both gases,the rate of diffusion is proportional to the mass $m$ diffused: $r = \frac{m}{t}$.
Therefore,$\frac{m_{H_2}}{m_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
Given $m_{H_2} = 2 \ g$,$M_{H_2} = 2 \ g/mol$,and $M_{O_2} = 32 \ g/mol$.
Substituting the values: $\frac{2}{m_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Thus,$m_{O_2} = \frac{2}{4} = 0.5 \ g$.

(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: $r_{CH_4} = 2 \times r_X$.
Using the ratio: $\frac{r_{CH_4}}{r_X} = \sqrt{\frac{M_X}{M_{CH_4}}}$.
Substituting the values: $2 = \sqrt{\frac{M_X}{16}}$.
Squaring both sides: $4 = \frac{M_X}{16}$.
Therefore,$M_X = 4 \times 16 = 64 \ g/mol$.

(B) According to Graham's law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Since $r = \frac{V}{t}$,for the same volume $V$,we have $\frac{V}{t} \propto \frac{1}{\sqrt{M}}$,which implies $t \propto \sqrt{M}$.
Therefore,the ratio of times is $\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$.
Given $t_{H_2} = 5 \ s$ and $M_{H_2} = 2 \ g/mol$:
For $He$ $(M = 4)$: $t = 5 \sqrt{\frac{4}{2}} = 5\sqrt{2} \ s \approx 7.07 \ s$.
For $O_2$ $(M = 32)$: $t = 5 \sqrt{\frac{32}{2}} = 5 \times 4 = 20 \ s$.
For $CO$ $(M = 28)$: $t = 5 \sqrt{\frac{28}{2}} = 5\sqrt{14} \ s \approx 18.7 \ s$.
For $CO_2$ $(M = 44)$: $t = 5 \sqrt{\frac{44}{2}} = 5\sqrt{22} \ s \approx 23.45 \ s$.
Thus,the correct match is $20 \ s$ for $O_2$.

(D) $Boyle$,$Charles$,and $Avogadro$ are famous for their fundamental gas laws ($Boyle's \ Law$,$Charles' \ Law$,and $Avogadro's \ Law$ respectively).
$Faraday$ is primarily known for his contributions to electromagnetism and electrochemistry,not for the development of gas laws.

(C) According to Dalton's Law of Partial Pressures,the total pressure $P$ in the vessel is the sum of the partial pressures of nitrogen $(P_{N_2})$ and oxygen $(P_{O_2})$.
Since the number of molecules of nitrogen and oxygen are equal,their partial pressures are equal: $P_{N_2} = P_{O_2} = P/2$.
If nitrogen is removed,only oxygen remains in the vessel.
Therefore,the new pressure will be equal to the partial pressure of oxygen,which is $P/2$.

(A) According to Graham's law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
For hydrogen gas $(H_2)$,$M_1 = 2 \ g/mol$.
For helium gas $(He)$,$M_2 = 4 \ g/mol$.
Substituting these values into the formula: $\frac{r_{H_2}}{r_{He}} = \sqrt{\frac{4}{2}} = \sqrt{2} \approx 1.414$.
Therefore,the rate of diffusion of hydrogen gas is approximately $1.4$ times that of $He$ gas.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given that the rate of diffusion of hydrogen $(r_H)$ is $6$ times the rate of diffusion of gas $A$ $(r_A)$,we have $r_H = 6r_A$.
The molar mass of hydrogen $(H_2)$ is $M_H = 2 \ g/mol$.
Substituting these values into the formula: $\frac{r_H}{r_A} = \sqrt{\frac{M_A}{M_H}}$.
$6 = \sqrt{\frac{M_A}{2}}$.
Squaring both sides: $36 = \frac{M_A}{2}$.
Therefore,$M_A = 36 \times 2 = 72 \ g/mol$.

(B) Inert gases are monoatomic gases.
For a monoatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$ and at constant pressure is $C_P = \frac{5}{2}R$.
The ratio $\gamma$ is given by $\gamma = \frac{C_P}{C_V} = \frac{5/2R}{3/2R} = \frac{5}{3} \approx 1.66$.

(D) For diatomic gases,the degrees of freedom $(f)$ change with temperature due to the activation of vibrational modes.
At lower temperatures,only translational and rotational degrees of freedom are active $(f = 5)$,so $\gamma = C_p/C_v = 1 + (2/f) = 1 + (2/5) = 1.40$.
At higher temperatures,vibrational modes also become active,increasing the degrees of freedom $(f = 7)$,so $\gamma = 1 + (2/7) \approx 1.29$.
The value $1.66$ corresponds to monoatomic gases.
Therefore,statements $1$ and $3$ are correct.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ and inversely proportional to the time $t$ taken for the same amount of gas to diffuse: $r \propto \frac{1}{\sqrt{M}}$ and $r \propto \frac{1}{t}$.
Therefore,$\frac{r_1}{r_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$.
Given $t_{O_2} = 18 \ sec$,$M_{O_2} = 32 \ g/mol$,$t_g = 45 \ sec$,and $M_g = ?$.
Substituting the values: $\frac{45}{18} = \sqrt{\frac{M_g}{32}}$.
Squaring both sides: $(\frac{45}{18})^2 = \frac{M_g}{32}$.
Thus,$M_g = 32 \times \frac{45^2}{18^2}$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
For $SO_2$ $(M = 64)$,$O_2$ $(M = 32)$,and $CH_4$ $(M = 16)$:
$r_{SO_2} : r_{O_2} : r_{CH_4} = \frac{1}{\sqrt{64}} : \frac{1}{\sqrt{32}} : \frac{1}{\sqrt{16}}$
$= \frac{1}{8} : \frac{1}{4\sqrt{2}} : \frac{1}{4}$
Multiplying the entire ratio by $8$ to simplify:
$= 1 : \frac{8}{4\sqrt{2}} : 2$
$= 1 : \frac{2}{\sqrt{2}} : 2$
$= 1 : \sqrt{2} : 2$

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
For two gases,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given $r_{H_2} = \frac{50 \ mL}{20 \ min} = 2.5 \ mL/min$ and $M_{H_2} = 2 \ g/mol$.
For oxygen,$M_{O_2} = 32 \ g/mol$ and volume $V = 40 \ mL$.
Let the time taken be $t$. Then $r_{O_2} = \frac{40}{t}$.
Substituting the values: $\frac{2.5}{40/t} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
$\frac{2.5 \times t}{40} = 4 \implies 2.5 \times t = 160$.
$t = \frac{160}{2.5} = 64 \ min$.

(C) According to Dalton's Law of Partial Pressures,the total pressure of a gas collected over water is the sum of the partial pressure of the dry gas and the vapour pressure of water at that temperature.
$P_{\text{total}} = P_{O_2} + P_{H_2O}$
Given:
$P_{\text{total}} = 751\, mm\, Hg$
$P_{H_2O} = 21\, mm\, Hg$
Therefore,$P_{O_2} = P_{\text{total}} - P_{H_2O} = 751 - 21 = 730\, mm\, Hg$.
To convert this to atmospheres,we use the relation $1\, atm = 760\, mm\, Hg$:
$P_{O_2} = \frac{730}{760}\, atm \approx 0.96\, atm$.

(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For helium $(He)$ and methane $(CH_4)$,the ratio is given by: $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{He}}}$.
The molar mass of methane $(CH_4)$ is $16 \ g/mol$ and helium $(He)$ is $4 \ g/mol$.
Substituting the values: $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.

(B) $ (b) $ The internal energy $(U)$ of an ideal gas is a function of temperature $(T)$ only.
For an ideal gas,there are no intermolecular forces of attraction,so the internal energy does not depend on volume or pressure.

(B) An $isochoric$ process is defined as a thermodynamic process that occurs at constant volume.
Since the volume remains constant,the change in volume is zero,i.e.,$dV = 0$.

(D) Mechanical work,defined as $w = -P_{ext} \Delta V$,is significant only in systems containing gases because they undergo appreciable changes in volume during expansion or compression. Solids and liquids exhibit negligible volume changes,making mechanical work in these systems effectively zero.

(A) For diatomic gases like $H_2$,the degrees of freedom $f = 5$.
The ratio of heat capacities is given by $\gamma = C_p/C_v = 1 + 2/f$.
Substituting $f = 5$,we get $\gamma = 1 + 2/5 = 1 + 0.4 = 1.4$.
Thus,the correct option is $A$.

(A) Solubility of a gas in water depends on its ability to react with water or its polarity.
$SO_2$ reacts with water to form sulfurous acid $(H_2SO_3)$,making it highly soluble.
$CO_2$ is polar and reacts slightly with water to form carbonic acid $(H_2CO_3)$,making it moderately soluble.
$O_2$ is non-polar but has some solubility due to induced dipole-induced dipole interactions.
$H_2$ is a non-polar,small molecule with very weak van der Waals forces,making it the least soluble among the given options.

(B) An oxygen-helium mixture is used for artificial respiration in deep sea diving instead of air.
This is because the nitrogen present in air dissolves in the blood under high pressure when a sea diver goes into the deep sea.
When the diver returns to the surface,the decrease in pressure causes nitrogen to form bubbles in the blood,which leads to severe pain.
This condition is known as "bends".

(A) Deep sea divers face the risk of 'the bends' (decompression sickness) due to the high solubility of nitrogen in blood under high pressure.
When the diver ascends,the pressure decreases,causing nitrogen to form bubbles in the blood,which can be fatal.
Helium is used as a diluent for oxygen because it is much less soluble in blood than nitrogen,even at high pressures,thereby preventing the formation of bubbles during decompression.

(C) For monoatomic gases,the degrees of freedom $(f)$ is $3$.
The molar heat capacity at constant volume is $C_v = \frac{f}{2}R = \frac{3}{2}R$.
The molar heat capacity at constant pressure is $C_p = C_v + R = \frac{3}{2}R + R = \frac{5}{2}R$.
The ratio $\gamma = \frac{C_p}{C_v} = \frac{5/2R}{3/2R} = \frac{5}{3} \approx 1.66$.
Since inert gases are monoatomic,the ratio is $1.66$.

(A) The marsh gas detector used by miners works on the principle of the difference in the rates of diffusion of gases.
Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass $(Rate \propto \frac{1}{\sqrt{M}})$.
Marsh gas,primarily composed of methane $(CH_4)$,diffuses at a different rate compared to air. This difference in diffusion rates allows the detector to identify the presence of methane in the mine atmosphere,ensuring safety for the miners.

(D) Ordinary evaporation is a surface phenomenon where some molecules have enough kinetic energy to escape into the vapour phase.
If the container is closed,an equilibrium is eventually reached where the rate of evaporation equals the rate of condensation.
The pressure exerted by the vapour at this state of equilibrium,where liquid and vapour coexist,is defined as the $Saturated \ vapour \ pressure$.

(B) The atmospheric pressure is defined as the sum of the partial pressures of all the gaseous constituents present in the atmosphere.
At sea level,this total pressure is defined as $1 \, atm = 760 \, mm \, Hg$ or $760 \, torr$.

(B) In a pressure cooker,the pressure inside is higher than the atmospheric pressure.
According to the principle of elevation of boiling point,as the pressure increases,the boiling point of water increases.
This allows the food to cook at a higher temperature,which significantly reduces the cooking time.

(B) At the $Boiling \ point$,the vapour pressure of a liquid becomes equal to the atmospheric pressure.
At this temperature,the liquid molecules gain enough kinetic energy to overcome the intermolecular forces and escape into the gaseous phase.

(B) During the process of evaporation,high-energy molecules escape from the surface of the liquid. As a result,the average kinetic energy of the remaining molecules decreases,which leads to a decrease in the temperature of the liquid.