Chemistry · Physics

Gases and Gas Laws

298 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 quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

Two flasks A and B of equal volumes are kept under similar conditions of temperature and pressure. If flask A holds 16.2 g of gas X while flask B holds 1.012 g of hydrogen, calculate the relative molecular mass of gas X:

  1. 20 g

  2. 32 g

  3. 28 g

  4. 44 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A mole is equal to the relative molecular mass of a gas. As given, the number of moles of gas X and hydrogen are same in the two flasks.
Therefore no. of moles in 2g of hydrogen gas = 1
No. of moles in 1.01g of hydrogen gas = 1/2
Now the weight of gas X that contains 1/2 moles = 16.2g
Weight of gas X that contains 1mole = $\frac{16.2}{1/2}$ = 32g

Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

The weight of $1$ litre of a glass at STP is $2$ grams, its molecular weight is:

  1. $44.4$
  2. $44.8$
  3. $44.1$
  4. $55.8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$At\quad STP,\quad volume\quad of\quad 1\quad mole\quad gas\quad =\quad 22.4\quad L\ Mass\quad of\quad 1L\quad gas\quad =\quad 2g\ Mass\quad of\quad 22.4L\quad gas\quad =\quad 2\times 22.4\quad =\quad 44.8\ So,\quad molecular\quad weight\quad of\quad gas\quad =\quad 44.8\ So,\quad correct\quad answer\quad is\quad option\quad B.$

Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

M g of a substance when vaporised occupy a volume of 5.6 litre at NTP. The molecular mass of the substance will be: 

  1. $M$
  2. $2M$
  3. $3M$
  4. $4M$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,
$Mg$ of substance occupy volume$=5.6\,litre $ at NTP.
At NTP,
1 mol occupy 22.4 litre of volume.
5.6 litre$=Mg$
22.4 litres$=4\,Mg$ of substance.
So, Molecular mass of gas$=4\,Mg/mol$
Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Gaseous $ N _{2}O _{4} $ dissociates into gaseous $ NO _{2} $ according to the reaction $ N _{2}O _{4} (g) \rightleftharpoons 2NO _{2}(g)$ at 300 K and 1 atm pressure, the degree of dissociation of $ N _{2}O _{4} $ is 0.2. If one mole of $ N _{2}O _{4} $ gas is contained in a vessel, then the density of the equilibrium mixture is : 

  1. 3.11 g/L

  2. 4.56 g/L

  3. 1.56 g/L

  4. 6.22 g/L
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$N _2O _4\longrightarrow 2NO _2$


at $t=0$, moles of $N _2O _4=1$, moles of $NO _2=0$

at $t=equilibrium$, mole of $N _2O _4=1-a$, mole of $NO _2=2a$

$a$=degree of dissociation.

Molecular weight of mixture$=\cfrac {(1-a)\times\text{molar mass of }N _2O _4+2a\times \text{molar mass of }NO _2}{(1-a+2a)}$
                                                $=\cfrac {(1-0.2)(28+64)+2\times 0.2\times (14+32)}{1+0.2}$
$M=76.66$

$P=1 atm,T=300K,$

$d=PM/RT$

    $=\cfrac {1\times 76.66}{0.082 \times 300}=3.11\  gm/lit\ $ .

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 } $
$\Rightarrow n=2$