Chemistry · Physics

Gases and Gas Laws

252 Questions

The study of gases and gas laws involves understanding the relationships between pressure, volume, and temperature of gases. This topic is essential for chemistry and physics sections in many competitive exams. Practice these questions to master concepts like the ideal gas law, partial pressure, and molecular properties.

Ideal gas law calculationsGas volume and pressureStoichiometry of gasesDensity of gasesBoltzmann constant applicationsThermal speed of sound

Gases and Gas Laws Questions

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

A mixture of CO and $CO _2$ has a density of $1.5 g/l $ at $27^o$C and $760$ mm pressure. If $1\ l$ of the mixture is exposed to alkali, what would be the pressure of the remaining gas at the same volume and temperature?

  1. $533$ mm
  2. $473$ mm
  3. $335$ mm
  4. $595$ mm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given density$=d=\dfrac{m}{volume}$ …….$(1)$


so the given mass is a mixture of CO & $CO _2$

$m=n _{CO}\times 28+n _{CO _2}\times 44$

${$ using no. of moles$=\dfrac{mass}{mol. mass}}$

Also using $PV=nRT$
$V=\dfrac{(n _{CO}+n _{CO _2})RT}{P}$

Putting in $(1)$
$1.5=\dfrac{n _{CO}\times 28+44\times n _{CO _2}}{{(n _{CO _2}+n _{CO})\times 0.082\times 300}}$ $[{760$mm$=1$atm$}, P=1atm]$

$1.5=\dfrac{28n _{CO}+44n _{CO _2}}{24.6(n _{CO}+n _{CO _2})}$


$\Rightarrow 36.9(n _{CO}+n _{CO _2})=28n _{CO}+44n _{CO _2}$

$9n _{CO}=7n _{CO _2}$

After the reaction with alkali, all $CO _2$ will be used so the remaining pressure will be of $CO$.

$P _{CO _2}=\dfrac{n _{CO}}{n _{CO}+n _{CO _2}}\times P _{Total}$

$=\dfrac{n _{CO}}{n _{CO}+\dfrac{9}{7}n _{CO}}$

$=\dfrac{7}{16}P _{Total}=\dfrac{7}{16}\times 760$

$ \approx 335$ mm
Option C.

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

A quantity of $4$g of oxygen occupies $10$ L at a particular pressure and temperature. If the pressure of gas is doubled and absolute temperature is halved, in order to maintain constant volume.

  1. $3$ g gas should be removed from container
  2. $3$ g gas should be added in the container
  3. $16$ g gas should be added in the container
  4. $12$ g gas should be added in the container
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,
$m _1=4g$ $V _1=10L$,$P _1=P$,$T _1=T$

$m _2=?$, $V _2=10L$,$P _2=2P$, $T _2=\dfrac{T}{2}$

We know that,

$PV=\dfrac{m}{M}RT$

At constant volume, V

$\dfrac{P _1}{P _2}=\dfrac{m _1RT _1}{m _2RT _2}$

$\dfrac{P}{2P}=\dfrac{4\times T}{m _2\times \dfrac{T}{2}}$


$\dfrac{1}{2}=\dfrac{8}{m _2}$

$m _2=16$g

Mass of gas added into container$=16-4=12$g.

Multiple choice geography in search of the source of wind variations in atmospheric pressure properties of air pressure belts the concept of bernoulli's principle air exerts pressure

What is the level of mercury at normal atmospheric pressure?

  1. 64 cm

  2. 70 cm

  3. 76 cm

  4. 80 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$76cm$

Atmospheric pressure is expressed in several different systems of units: millimetres (or inches) of mercury, pounds per square inch (psi), dynes per square centimetre, millibars (mb), standard atmospheres, or kilopascals. Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, 14.70 pounds per square inch, 1,013.25 × 103 dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals. Variations about these values are quite small; for example, the highest and lowest sea-level pressures ever recorded are 32.01 inches (in the middle of Siberia) and 25.90 inches (in a typhoon in the South Pacific). The small variations in pressure that do exist largely determine the wind and storm patterns of Earth.

Multiple choice chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

Collision frequency of a gas at $1\ atm$ pressure is $Z$. Its value at $0.5\ atm$ will be:

  1. $0.25Z$
  2. $2Z$
  3. $0.50Z$
  4. $Z$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\rightarrow$ As the pressure increases collision frequency increases.

$\rightarrow$ Vice versa pressure is decreased to half, frequency of collision decreases to half.
Hence option $C$ is correct.

Multiple choice chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

One mole of helium and one mole of neon are taken in a vessel. Which of the following statements are correct?

  1. Molecules of helium strike the wall of vessel more frequently

  2. Moles of neon apply more average force per collision on the wall of vessel

  3. Molecules of helium have greater average molecular speed

  4. Helium exerts higher pressure than neon
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Helium (molar mass 4) has a higher average speed than Neon (molar mass 20) at the same temperature. Because collision frequency with walls is proportional to average speed, Helium molecules strike the walls more frequently.

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

A gas has a molecular diameter of 0.1 m. It also has a mean free path of 2.25 m. What is its density?

  1. $10^{-3}$
  2. $10^{-2}$
  3. $10^{-4}$
  4. $10^{-5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given :
Molecular diameter of a gas, $d = 0.1 m$.
Mean free path, $l = 2.25 m$.
The mean free path traversed by the molecules is given by
$l = \dfrac{1}{\sqrt 2 \pi d^2 \rho}$
Therefor,
$\rho = \dfrac{1}{\sqrt 2 \pi d^2 l}$
Using the given values we get,
$\rho = \dfrac{1}{\sqrt 2 (3.14)(0.1)^2 (2.25)}$
$\rho = \dfrac{1}{0.0999}$
$\rho = 10 ^{-3}$
Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

Estimate the mean free path of nitrogen molecule in a cylinder containing nitrogen at 2.0atm pressure and temperature ${17^o}C$.(take the radius of nitrogen molecule to be 1.0A, Molecular mass=28gm

  1. $2.25x{10^{ - 8}}m$
  2. $1.12x{10^{ - 7}}m$
  3. $11.2x{10^{ - 7}}m$
  4. $22.5x{10^{ - 8}}m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the ideal gas law to find number density n = P/kT, then applying the mean free path formula lambda = 1 / (sqrt(2) * pi * n * d^2) with d = 2 * 10^-10 m, we get approximately 1.12 * 10^-7 m.

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

Modern vacuum pumps can evacuate a vessel down to a pressure of $4.0\times { 10 }^{ -15 }atm$. At room temperature $(300K)$, taking $R=8.3J{ K }^{ -1 }\quad { mole }^{ -1 },1\quad atm={ 10 }^{ 5 }Pa\quad \quad $ and ${ N } _{ Avagadro }=6\times { 10 }^{ 23 }{ mole }^{ -1 }$, the mean distance between the molecules of gas in an evacuated vessel will be of the order of :

  1. $0.2\mu$ $m$
  2. $0.3\mu$ $m$
  3. $0.2$mm
  4. $0.2nm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
As we know formula for mean free path
$Y=\dfrac{KT}{\sqrt{2}\pi{\sigma}^{2}p}$
where $\sigma=$diameter of the molecule
$p=$pressure of the gas
$T=$Temperature
$K=$Boltzmann's constant.
Let intermolecular distance be $D$ then in a volume $\dfrac{4\pi}{3}{D}^{3}$ there is only one
$\dfrac{4\pi}{3}{D}^{3}p=\dfrac{1}{{N} _{A}}={R} _{T}$
or $D={\left(\dfrac{3RT}{4\pi{N} _{A}p}\right)}^{\frac{1}{3}}$
Put $p=4\times{10}^{-10}$Pa
$R=83$,${N} _{A}=6\times{10}^{23}$ and $T=300$K
$D={\left(\dfrac{3\times 83 \times 300}{4\times\dfrac{22}{7}\times 6\times{10}^{23}\times 4\times{10}^{-10}}\right)}^{\frac{1}{3}}$
$=0.2$mm
Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

The mean free path of the molecule of a certain gas at 300 K is $2.6\times10^{-5}:m$. The collision diameter of the molecule is 0.26 nm. Calculate
(a) pressure of the gas, and
(b) number of molecules per unit volume of the gas.

  1. (a) $1.281\times 10^{23}\:m^{-3}$ (b) $5.306\times 10^{2}\:Pa$
  2. (a) $1.281\times 10^{22}\:m^{-3}$ (b) $5.306\times 10^{3}\:Pa$
  3. (a) $12.81\times 10^{23}\:m^{-3}$ (b) $53.06\times 10^{2}\:Pa$
  4. (a) $2.56\times 10^{23}\:m^{-3}$ (b) $10.612\times 10^{2}\:Pa$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \lambda =2.6\times 10^{-5}:m, :\sigma =0.26:nm=2.6\times 10^{-10}m$
$\displaystyle T=300:K$
$\displaystyle \lambda =\frac{1}{\sqrt{2}\pi \sigma ^{2}N^{\ast }}$
$\displaystyle 2.6\times 10^{-5}=\frac{1}{\sqrt{2}\times 3.14\times (2.6\times 10^{-10})^{2}\times N^{\ast }}$
$\displaystyle N^{\ast }=1.281\times 10^{23}m^{-3}$
$\displaystyle N^{\ast }=\frac{P}{KT}$
$\displaystyle P=1.281\times 10^{23}\times 1.38\times 10^{-23}\times 300$
$\displaystyle P=530.3:Pa$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

A gas has a density of $10$ particles$/m^3$ and a molecular diameter of $0.1 $m. What is its mean free path?

  1. $2.25m$
  2. $1m$
  3. $3m$
  4. $0.25m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mean free path estimated by the kinetic theory of gases is given by

$\lambda = \displaystyle \frac{1}{\sqrt{2}\pi nd^2}$
Given that number density  $n = 10\textrm{ m}^{-3}$ and diameter $d = 0.1\textrm { m}$
Thus, $\lambda = \displaystyle \frac{1}{\sqrt{2}\pi \times 10\times 0.1^2} \approx 2.25 \textrm{ m}$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

A gas in a 1 $m^3$ container has a molecular diameter of 0.1 m. There are 10 molecules. What is its mean free path?

  1. 2.25m

  2. 2m

  3. 3m

  4. 1m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mean free path of a molecule is given by the formula ($\lambda $)  $\cfrac{1}{\sqrt{2\pi d^2 n}}$ 

where d is the diameter of the molecules; n - number of molecules

$\cfrac{1}{\sqrt{2\pi \times 0.1 \times 0.1 \times 10}} = 2.25m$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

A container is divided into two equal parts I and II by a partition with a small hole of diameter d. The two partitions are filled with same ideal gas, but held at temperatures $T _I=150$K and $T _{II}=300$K by connecting to heat reservoirs. Let $\lambda _I$ and $\lambda _{II}$ be the mean free paths of the gas particles in the two parts such that $d > > \lambda _I$ and $d > > \lambda _{II}$. Then $\lambda _I/\lambda _{II}$ is close to.

  1. $0.25$
  2. $0.5$
  3. $0.7$
  4. $1.0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

Partition has hole of diameter $d$, Mean pressure between both sections is equal.

Boltzmann constant ${{K} _{B}}$

At constant pressure, Mean Free path $\lambda \ \alpha \ \sqrt{{{K} _{B}}T}$

Mean free path in ${{1}^{st}}$ section ${{\lambda } _{I}}=\sqrt{{{K} _{B}}\times 150}$

Mean free path in ${{2}^{nd}}$ section ${{\lambda } _{II}}=\sqrt{{{K} _{B}}\times 300}$

$\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}=\dfrac{\sqrt{{{K} _{B}}\times 150}}{\sqrt{{{K} _{B}}\times 300}}=0.707$

Hence, $\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}\cong 0.7$ 

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

Calculate the means free path of nitrogen molecule at $27^o$C when pressure is $1.0$ atm. Given, diameter of nitrogen molecule $=1.5\overset{o}{A}$, $k _B=1.38\times 10^{-23}$J $K^{-1}$. If the average speed of nitrogen molecule is $675$ $ms^{-1}$. The time taken by the molecule between two successive collisions is?

  1. $0.6$ns
  2. $0.4$ns
  3. $0.8$ns
  4. $0.3$ns
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here, $T=27^oC=27+273=300$K.
$P=1$atm $=1.01\times 10^5$N $m^{-2}$, d$=1.5\overset{o}{A}=1.5\times 10^{-10}$m,
$k _B=1.38\times 10^{-23}$J $K^{-1}, \lambda =?$
From $\lambda =\dfrac{k _BT}{\sqrt{2}\pi d^2p}=\dfrac{1.38\times 10^{-23}\times 300}{1.414\times 3.14(1.5\times 10^{-10})^2\times 1.01\times 10^5}$
$=4.1\times 10^{-7}$m
Time interval between two successive collisions
$t=\dfrac{distance}{speed}=\dfrac{\lambda}{v} = \dfrac{4.1\times 10^{-7}}{675}=0.6\times 10^{-9}s$