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 chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

At 373 K,a gaseous reaction $A\rightarrow 2B+C$ is found to be of first order.Starting with pure A,the total pressure at the end of 10 min was 176 mm of Hg and after a long time when A was completely dissociated,it was 270 mm of Hg.The pressure of A at the end of  10 minutes was:

  1. 94 mm of Hg

  2. 47 mm of Hg

  3. 43 mm of Hg

  4. 90 mm of Hg
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

                    $A \to 2B + C$

$at\,t = 0$   $x$       $0$    $0$

$at\,t = 10$    $x-y$           $2y$    $y$  

 $total=x-y+2y+y$

        $=x+2y=176\,mm$-------$(i)$

$at\,{t={100}}$   $0$       $2x$    $x$

$total=2x+x=270\,mm$

$ \Rightarrow 3x = 270$
$ \Rightarrow x = 90\,mm\,\,of\,Hg$

put the value of $A$ in $e{q^n}\,(i),$ we get
   $x+2y=176$
$ \Rightarrow 90 + 2y = 176$
$ \Rightarrow  2y = 86$
$ \Rightarrow y=43\,\,\,mm\,\,of\,Hg$

At the end of $10$ min pessure of $A$ is  $x-y=90-43=47\,mm\,of\,Hg$

Option B is correct.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

A $10\ litre$ box contains $O _3$ and $O _2$ at equilibrium at 2000 K. $K _p=4 \times 10^{14}$ atm for $2O _3(g) \rightleftharpoons  3O _2(g)$. Assume that $P _{O _2} > > P _{O _3}$ and if total pressure is 8 atm, then patial pressure of $O _3$ will be: 

  1. $8 \times 10^{-5} atm$
  2. $11.3 \times 10^{-7} atm$
  3. $9.71 \times 10^{-6} atm$
  4. $8 \times 10^{-2} atm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Kp = P_O2^3 / P_O3^2 = 4 * 10^14. Total pressure = P_O2 + P_O3 = 8. Since P_O2 >> P_O3, P_O2 approx 8. 8^3 / P_O3^2 = 4 * 10^14. 512 / P_O3^2 = 4 * 10^14. P_O3^2 = 128 * 10^-14. P_O3 = sqrt(128) * 10^-7 = 11.3 * 10^-7.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Identical cylinders contain helium at 2.5 atm and agron at 1 atm respectively. If the are filled  in one of the cylinder,the pressure would be.

  1. 3.6 atm

  2. 1.75 atm

  3. 1.5 atm

  4. 1.0 atm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that,

Pressure ${{p} _{1}}=2.5\,atm$

Pressure ${{P} _{2}}=1\,atm$

Volume ${{V} _{1}}={{V} _{2}}=V$

Both the cylinders are similar, volume of both the gases is equal

Let the volume of gases be V

The amount of pressure P in one of the cylinder will be equal to the total pressure at equilibrium

Now,

  $ P=\dfrac{{{P} _{1}}{{V} _{1}}+{{P} _{2}}{{V} _{2}}}{{{V} _{1}}+{{V} _{2}}} $

 $ P=\dfrac{2.5\times V+1\times V}{2V} $

 $ P=1.75\ atm $

Hence, the pressure is $1.75\ atm$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

If a given mass of gas occupies a volume of 10 cc at 1 atmospheric pressure and temperature 100$^o$C. What will be its volume at 4 atmospheric pressure, the temperature being the same?

  1. 100 cc

  2. 400 cc

  3. 1.04 cc

  4. 2.5 cc
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

It is an isothermal process.
$P _1 V _1 = P _2 V _2$
$1 \times 10 = 4 \times V _2$
$V _2 = 2.5 cc$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

A thin tube of uniform cross-section is sealed at both ends. When it lies horizontally, the middle $5$cm length contains mercury and the two equal ends contain air at the same pressure P. When the tube is held at an angle of $60^o$ with the vertical, then the lengths of the air columns above and below the mercury column are $46$cm and $44.5$cm respectively. Calculate the pressure P in cm of mercury. The temperature of the system is kept at $30^o$C.

  1. $75.4$
  2. $45.8$
  3. $67.5$
  4. $89.3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let A be the area of cross-section of the tube. When the tube is horizontal, the $5$cm column of Hg is in the middle, so length of air column on either side at pressure $P=\dfrac{46+44.5}{2}=45.25$cm
When the tube is held at $60^o$ with the vertical, the lengths of air columns at the bottom and the top are $44.5$cm and $46$cm respectively. If $P _1$ and $P _2$ are their pressures, then $P _1-P _2=5\cos 60^o=5\times \dfrac{1}{2}=\dfrac{5}{2}$cm of Hg
Using Boyle's law for constant temperature,
$PV=P _1V _1=P _2V _2$
$P\times A\times 45.25=P _1\times A\times 44.5=P _2\times A\times 46$
$\therefore \dfrac{P\times 45.25}{44.5}-\dfrac{P\times 45.25}{46}=\dfrac{5}{2}$
or $P=\dfrac{5\times 44.5\times 46}{2\times 45.25\times 1.5}=75.4$cm of Hg

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

The density of air at NTP is $1.293\space kgm^{-3}$ and density of mercury at $0^{\small\circ}\space C$ is $13.6\times10^3 \space kgm^{-3}$. If $C _p = 0.2417\space calkg^{-10}C^{-1}$ and $C _v = 0.1715$, the speed of sound in air at $100^{\small\circ}\space C$ will be $(g = 9.8\space Nkg^{-1})$

  1. $260\space ms^{-1}$
  2. $332\space ms^{-1}$
  3. $350.2\space ms^{-1}$
  4. $369.4\space ms^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
NTP conditions:        $T = 25  ^o C=  298.15  K                P =  1  bar  =  10^5    Pa$

Given:   Density of air at NTP  $\rho = 1.293     kg /m^3$

$\gamma =  \dfrac{C _p}{C _v} = \dfrac{0.2417}{0.1715} = 1.4$

Speed of sound in air at NTP,      $v _{25^o C} =  \sqrt{\dfrac{\gamma  P}{\rho} }  = \sqrt{\dfrac{1.4  \times 10^5}{1.293}}  = 330.15   m/s$

Let speed of sound in air at $100^o  C$ be  $v _{100^o  C}$

As     $v   \propto  \sqrt{T}$


Thus   $\dfrac{v _{100^o  C}}{v _{25^o  C} } = \sqrt{\dfrac{373.15}{298.15}} = 1.118$

$\implies  v _{100^o  C} = 1.118 \times  330.15 = 369.35   m/s$

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

The speed of sound in hydrogen at $  N T P,  $ is 1270 $ \mathrm{m} / \mathrm{s} .$ Then the speed in a mixture of hydrogen and oxigen in the ratio $  4 : 1  $ by volume, (in $  m / s )  $ will be

  1. 635

  2. 318

  3. 158

  4. 1270
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Speed of sound v = sqrt(gamma * R * T / M). For a mixture, M_mix = (n1M1 + n2M2) / (n1 + n2). With 4:1 ratio, M_mix = (4*2 + 1*32) / 5 = 40/5 = 8. Since v is inversely proportional to sqrt(M), v_mix = v_H2 * sqrt(M_H2 / M_mix) = 1270 * sqrt(2 / 8) = 1270 * 0.5 = 635 m/s.

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

Two moles of hydrogen are mixed with n moles of helium. The root mean square speed of gas molecules in the mixture is $\sqrt2$ times the speed of sound in the mixture. Then n is 

  1. $3$
  2. $2$
  3. $1.5$
  4. $2.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

v_rms = sqrt(3RT/M_mix). v_sound = sqrt(gamma_mix * RT / M_mix). Given v_rms = sqrt(2) * v_sound, then 3RT/M_mix = 2 * gamma_mix * RT / M_mix, so gamma_mix = 1.5. For a mixture, gamma = (n1Cp1 + n2Cp2) / (n1Cv1 + n2Cv2). With 2 moles H2 (gamma=1.4, Cv=2.5R) and n moles He (gamma=1.67, Cv=1.5R), solving for n yields 2.

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

Two moles of helium are mixed with $n$ moles of hydrogen. The root mean square $\left( rms \right) $ speed of gas molecules in the mixture is $\sqrt { 2 } $ times the speed of sound in the mixture. Then, the value of $n$ is

  1. $1$
  2. $3$
  3. $2$
  4. ${ 3 }/{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\because { v } _{ rms }=\sqrt { \dfrac { 3RT }{ M }  } $ and ${ v } _{ sound }=\sqrt { \dfrac { \gamma RT }{ M }  } $,
${ v } _{ rms }=2{ v } _{ sound }$
i.e. $\gamma =\dfrac { 3 }{ 2 } =$ ratio of $\dfrac { { C } _{ p } }{ { C } _{ V } } $ for the mixture
${ C } _{ V }=\dfrac { { n } _{ 1 }{ C } _{ { V } _{ 1 } }+{ n } _{ 2 }{ C } _{ { V } _{ 2 } } }{ { n } _{ 1 }+{ n } _{ 2 } } $
and ${ C } _{ p }=\dfrac { { n } _{ 1 }{ c } _{ { p } _{ 1 } }+{ n } _{ 2 }{ C } _{ { p } _{ 2 } } }{ { n } _{ 1 }+{ n } _{ 2 } } $
$\therefore \gamma =\dfrac { { C } _{ p } }{ { C } _{ V } } =\dfrac { { n } _{ 1 }{ C } _{ { p } _{ 1 } }+{ n } _{ 2 }{ C } _{ { p } _{ 2 } } }{ { n } _{ 1 }{ C } _{ { V } _{ 1 } }+{ n } _{ 2 }{ C } _{ { V } _{ 2 } } } $
$\therefore \dfrac { 3 }{ 2 } =\dfrac { 2\left( \dfrac { 5 }{ 2 } R \right) +n\left( \dfrac { 7 }{ 2 } R \right)  }{ 2\left( \dfrac { 3 }{ 2 } R \right) +n\left( \dfrac { 5 }{ 2 } R \right)  } $
$\Rightarrow \dfrac { 3 }{ 2 } =\dfrac { 10+7n }{ 6+5n } $
$\Rightarrow n=2$

Multiple choice physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

A liter of air at $20^oC$ is heated until both the pressure and the volume are tripled, what is the tempertare then.

  1. $2637^oK$
  2. $927^oK$
  3. $200^oK$
  4. $977^oK$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Applying the formula

$PV=nRT$
$\dfrac { { P } _{ 1 } }{ { V } _{ 1 } } =\dfrac { { P } _{ 2 } }{ { V } _{ 2 } } =\dfrac { { T } _{ 1 } }{ { T } _{ 2 } } $    [$R$ is constant]
Let ${ P } _{ 1 }=P$  and ${ V } _{ 1 }=V$
As given ${ P } _{ 2 }=3P$   ${ V } _{ 2 }=3V$
${ T } _{ 1 }={ 20 }^{ 0 }C=20+2+3=293$
${ T } _{ 2 }=?$
$\dfrac { PV }{ 3P\times 3V } =\dfrac { 293 }{ { T } _{ 2 } } $
$\dfrac { 1 }{ 9 } =\dfrac { 293 }{ { T } _{ 2 } } $
${ T } _{ 2 }=293\times 9=2637$
$\therefore$    New temperature $=2637$.

Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

1 litre of oxygen at a pressure of 1 atmosphere and 2 litres of nitrogen at a pressure of 0.5 atmosphere are introduced in a vessel of 1 litre capacity without any change in temperature. The total pressure in atmosphere is

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In this case, Initial pressure of oxygen $=$ 1 atm and initial volume of oxygen $=$ 1 lit
Initial pressure of nitrogen $=$ 0.5 atm and inititial volume of nitrogen $=$ 2 lit
Since temperature is constant, pressure (P) and volume (V) of the combined gas will be
$PV = P _{oxygen}V _{oxygen} + P _{nitrogen}V _{nitrogen}$
Given, volume of combined gas $=$ 1 lit.
${P}{(1)}={(1)}{(1)}+{(0.5)}{(2)}$
${P} ={2}$ atm
Putting all these values we get pressure of the combined gas $=$ 2 atm

Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

An ideal gas is trapped between Hg thread of $12\ cm$ and the closed lower end of a narrow vertical tube of uniform cross section. Length of the air column is $20.5\ cm$, when the open end is kept upward. If the tube is making $30^{0}$with the horizontal then the length of the air column is (assuming temperature to be constant and atmospheric pressure = $76\ cm$ of $Hg$) 

  1. $22\ cm$
  2. $18\ cm
  3. $24\ cm$
  4. $20.2\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Assuming the process to be isothermal, we can sa for an ideal gas,


${ P } _{ 1 }{ V } _{ 1 }={ P } _{ 2 }{ V } _{ 2 }$

$(76+12)(20.5)=(76+12sin({ 30 }^{ 0 }))l$, since in second case, the vertical height of mercury column will be $12sin({ 30 }^{ 0 })$.

This gives $l=22cm$.

Multiple choice common laboratory equipments common laboratory apparatus and equipments laboratory equipments know about some common gases chemistry

Two identical containers are filled with gas. The first container is filled neon and the second with krypton. Both containers are placed at NTP conditions.
Which of the following factors will NOT be equal for the two containers?

  1. Number of particles

  2. Kinetic Energy

  3. Pressure

  4. Masses

  5. Number of moles
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two identical containers are when filled with gas, one with $Ne$ and one with $Kr$ and are then placed at $NTP$. They will be present at same conditions of temperature and pressure.

Also, their volume will be same, since they occupy identical containers but their number of particles would be different.