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 physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

A vessel contains 1 mole of an ideal monoatomic gas. The coefficient of volume expansion of the gas is $\alpha $. 2 moles of a diatmoic; ideal gas is then introduced into the same vessel. The coefficient of the volume expansion of the mixture will be

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

The volume coefficient of gas is given by,
${ \alpha  } _{ V }={ \left( \frac { 1 }{ V } \frac { \partial V }{ \partial T }  \right)  } _{ p }$
From the above equation it can be seen that it is independent of the number of moles,

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

If at $60^\circ$C and 80 cm of mercury pressure, a definite mass of a gas is compressed slowly, then the final pressure of the gas if the final volume is half of the initial volume $ (\gamma = \dfrac { 3 }{ 2 }$) is:

  1. 120 cm of Hg

  2. 140 cm of Hg

  3. 160 cm of Hg

  4. 180 cm of Hg

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given initial pressure, $P _1=80\,cm\,of\,Hg$

If the gas is compressed slowly, then the process is isothermal.

At constant temperature,

$P _1V _1=P _2V _2$

Given, 

Final volume is half of the initial volume.

That is, $V _2=\dfrac{V _1}{2}$

Final pressure, $P _2=\dfrac{P _1V _1}{V _2}=\dfrac{80 \times V _1}{V _1/2}=160\,cm\,of\,Hg$
Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

A given amount of gas occupies 1000cc at 27$^{0}$ and 1200cc and 87$^{0}$ c. What is its volume  coefficient of expansion

  1. $\frac{1}{273}^{0}C^{-1}$
  2. $\frac{1}{173}^{0}C^{-1}$
  3. $173^{0}C^{-1}$
  4. $273^{0}C^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know , $\alpha =\frac { { V } _{ 2 }-{ V } _{ 1 } }{ { V } _{ 1 }{ t } _{ 2 }-{ V } _{ 2 }{ t } _{ 1 } } $
Substituting the values ${ V } _{ 2 }=1200cc$ , ${ V } _{ 1 }=1000cc$, ${ t } _{ 2 }={ 87 }^{ \circ  }C$, ${ t } _{1}={ 27 }^{ \circ  }C$.
$\therefore \alpha =\frac { 200 }{ \left( 87000-32400 \right)  } $
$\therefore \alpha ={ \frac { 1 }{ 273 }  }^{ \circ  }{ C }^{ -1 }$

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

$1$ mole of a gas with $\gamma =\dfrac{7}{5}$ is mixed with $1$ mole of gas with $\gamma =\dfrac{5}{3}$, the value of $\gamma$ of the resulting mixture of.

  1. $\dfrac{7}{5}$
  2. $\dfrac{2}{5}$
  3. $\dfrac{3}{2}$
  4. $\dfrac{12}{7}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ Y } _{ mis }=\cfrac { { n } _{ 1 }C{ \rho  } _{ 1 }+{ n } _{ 2 }C{ \rho  } _{ 2 } }{ { n } _{ 1 }C{ \gamma  } _{ 1 }+{ n } _{ 2 }C{ \gamma  } _{ 2 } } $

${ C\rho  } _{ 1 }=\cfrac { 5 }{ 2 } R$ then its $C{ v } _{ 1 }=\cfrac { 3 }{ 2 } R$
Because ${ C } _{ \rho  }-{ C } _{ v }=R$
for diatomic gas ${ C\rho  } _{ 2 }=\cfrac { 7R }{ 2 } $ then ${ Cv } _{ 2 }=\cfrac { 5 }{ 2 } R$
${ Y } _{ mis }=\cfrac { { n } _{ 1 }\times \cfrac { 5 }{ 2 } R+{ n } _{ 2 }\times \cfrac { 7 }{ 2 } R }{ { n } _{ 1 }\times \cfrac { 3 }{ 2 } R+{ n } _{ 2 }\times \cfrac { 5 }{ 2 } R } $
Here ${ n } _{ 1 }={ n } _{ 2 }=1$
${ Y } _{ mis }=\cfrac { 3 }{ 2 } $

Multiple choice physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

A mixture of ideal gases 7 kg of nitrogen and 11 Kg of $ CO _2 $ then (Take $\gamma$ for nitrogen and $CO _2$ as 1.4 and 1.3 respectively)

  1. Equivalent molecular weight of the mixture is 36.

  2. Equivalent molecular weight of the mixture is 18.

  3. $ \gamma $ for the mixture is 5/2
  4. $ \gamma $ for the mixture is 47/35
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Multiple choice physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

$3$ mole of gas ''X"  and $2$ moles of gas "Y" enters from end "P" and "Q" of the cylinder respectively. The cylinder has the area of cross section , shown as
under 
The length of the cylinder is $150cm$. The gas "X" intermixes with gas "Y" at the point . If the molecular weight of the gases X and Y is $20$ and $80$ respectively, then what will be the distance of point A from Q?

  1. $75cm$
  2. $50cm$
  3. $37.5$
  4. $90cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} \frac { { rx } }{ { ry } } =\frac { { { w _{ x } } } }{ { { n _{ y } } } } \sqrt { \frac { { { M _{ y } } } }{ { { M _{ x } } } }  }  \ =\frac { 3 }{ 2 } \sqrt { \frac { { 80 } }{ { 20 } }  } =\frac { 3 }{ 1 } =3:1 \ \therefore \frac { { dis\tan  ce\, \, travelled\, \, by\, \, gas\, \, X } }{ { dis\tan  ce\, \, travelled\, \, by\, \, gas\, \, Y } } =3:1 \ \therefore dis\tan  ce\, \, of\, \, A\, \, from\, \, Q=\frac { { 150 } }{ 3 } =50\, \, cms \end{array}$

Hence, OPtion $B$ is correct.

Multiple choice physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

The lowest pressure(the best Vaccum) that can be created in laboratory at 27 degree is $10^{-11} $ mm of Hg. At this pressure, the number of ideal gass molecules per $cm^{3}$ will be

  1. $3.22 \times 10 ^{12} $
  2. $1.61 \times 10 ^{12} $
  3. $3.21 \times 10 ^{6} $
  4. $3.22 \times 10 ^{5} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Use the ideal gas law PV = NkT, where N/V = P / (kT). Convert pressure to Pascals (1 mm Hg = 133.322 Pa) and temperature to Kelvin (300K). Calculate the number density N/V.

Multiple choice physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

One mole of gas occupies 10 ml at 50 mm pressure. The volume of 3 moles of the gas at 100 mm pressure and same temperature is 

  1. 15 ml

  2. 100 ml

  3. 200 ml

  4. 500 ml

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the ideal gas law PV = nRT, since T is constant, P1V1 / n1 = P2V2 / n2. Given P1=50, V1=10, n1=1; P2=100, n2=3. Solving for V2: (50 * 10) / 1 = (100 * V2) / 3. 500 = (100 * V2) / 3, so V2 = 1500 / 100 = 15 ml.

Multiple choice physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

The molecular weights of $O _2$ and $N _2$ are 32 and 28 respectively. At $15^0$C, the pressure of 1 gm will be the same as that of 1 gm in the same bottle at the temperature.

  1. $-21^0$C
  2. $13^0$C
  3. $15^0$C
  4. $56.4^0$C
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a fixed mass of gas in a constant volume, P is proportional to T/M. P1 = P2 implies T1/M1 = T2/M2. T1 = 15C = 288K. M1 = 32 (O2). M2 = 28 (N2). 288/32 = T2/28. 9 = T2/28. T2 = 252K. In Celsius, 252 - 273 = -21C.

Multiple choice physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

A vessel contains a mixture consisting of m$ _{1}$ - 7 g of nitrogen (M$ _{1}$ = 28) and m$ _{2}$ = 11 g of carbon dioxide (M$ _{2}$ = 44) at temperature T - 300 K and pressure P$ _{0}$ = 1 atm. The density of the mixture is

  1. $1.46g\ per\ litre$
  2. $2.567 g \ per \ litre$
  3. $3.752 g \ per \ litre$
  4. $4.572 g \ per \ litre$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the volume occupied $=V$

By Dalton's law of partial pressure
$\cfrac{P _{nit}}{P _0}=\cfrac{n _{nit}}{n _{nit}+n _{carb}}$
No. of moles of Nitrogen $\eta _{nit}=\cfrac{M _1}{M _{nit}}=\cfrac{7}{28}=0.25 mol$
No. of moles of carbon $\eta _{carb}=\cfrac{M _2}{M _{carb}}=\cfrac{11}{4}=0.25 mol$
Thus,
$P _{nit}=P _0\times\cfrac{0.25}{0.25\times0.25}=P _0/2=0.5atm$
From ideal gas equation
$V=\cfrac{nRT}{P}=\cfrac{0.25\times8.314\times290}{0.5\times101325}=0.0119mole$
Total mixture $m=(7+11)\times 10^{-11}kg$
Thus density s $P=\cfrac{m}{V}=\cfrac{(7+11)\times 10^{-3}}{0.0119}\approx1.46kg/m^3$
Option A is correct.

Multiple choice physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

A vessel of volume V contains a mixture of $1$mole of hydrogen and $1$ mole of oxygen(both considered as ideal). Let $f _1(v)dv$ denote the fraction of molecules with speed between v and $(v+dv)$ with $f _2(v)dv$, similarly for oxygen. then

  1. $f _1(v)+f _2(v)=f(v)$ obeys the Maxwell's distribution law
  2. $f _1(v), f _2(v)$ will obey the Maxwell's distribution law separately
  3. Neither $f _1(v)$ nor $f _2(v)$ will obey the Maxwell's distribution law
  4. $f _2(v)$ and $f _1(v)$ will be the same
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Maxwell-Boltzmann speed distribution function $\left(N _v=\dfrac{dN}{dv}\right)$ depends on the mass of the gas molecule. [Here, dN is the number of molecules with speeds between v and $(v+dv)$]. The masses of hydrogen and oxygen molecules are different.

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Two volumes of ammonia, on dissociation gave one volume of nitrogen and three volumes of hydrogen. How much hydrogen will be obtained from dissociation of $40 \,mL\, of\, NH _3$ ? 

  1. 60

  2. 40

  3. 80

  4. 50

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Its easy as $NH _3 : N _2 : H _2$ IS 2:1:3 As $NH _3$ is 40 ml hence form 60 ml of h2 and 20 ml of n2

so the answer is A

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

It was found from the chemical analysis of a gas that it has two hydrogen atoms for each carbon atoms. At $0^o C$ and $1\ atm$, its density is $1.25\ g$ per litre. The formula of the gas would be ______________.

  1. $CH _2$
  2. $C _2H _4$
  3. $C _2H _6$
  4. $C _4H _8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$PV=nRT$


$P=\dfrac {dRT}{M}$

$M=\dfrac {1.25\times 0.082\times 273}{1}$

$=27.98$

Molar mass $\equiv 28\ gram/mole$

$\therefore \ $ as empirical formula $=CH _2$

$\therefore \ $ emprical mass $=12+2=14$

$n=\dfrac {28}{14}=2$

$\therefore \ $ Molecular formula $=(CH _2) _2$

$=C _2H _4$

Hence, the correct option is $\text{B}$

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

When burnt in air, $14.0\ g$ mixture of carbon and sulpher gives a mixture of $CO 2$ and $SO _2$ in the volume ratio of $2:1$, volume being measured at the same conditions of temperature and pressure. Moles of carbon in the mixture is _________.

  1. $0.25$
  2. $0.40$
  3. $0.5$
  4. $0.75$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the mass of $C=x$


mass of $S=14-x$

moles of $C=\dfrac {x}{12};$ moles of $\dfrac {14-x}{32}$

$C+O _2\to CO _2;\quad S+O _2\to SO _2$

moles of $C=$ moles of $CO _2=\dfrac {x}{12}\quad $ moles of $S=$ moles of $SO _2=\dfrac {14-x}{32}$

Given,

$\dfrac {V _c}{V _s}=\dfrac {2}{1}$

$V\alpha $ no. of mole

$\dfrac {\dfrac {x}{12}}{\dfrac {14-x}{32}}=\dfrac {2}{1}$

$\dfrac {x}{12}\times \dfrac {32}{14-x}=\dfrac {2}{1}$

$\dfrac {4x}{42-3x}=1$

$4x=42-3x$

$7x=42$ 

$\boxed {x=6}$

$\therefore \ $ moles of $C$ in mixture $=\dfrac {w}{M}=\dfrac {6}{12}=0.5$

Answer is option $C$