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

For $10$ min each, at $27^o$C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of $3$l capacity. The resulting pressure is $4.18$ bar and the mixture contains $0.4$ mole of nitrogen. The molar mass of the unknown gas is?

  1. $112$g $mol^{-1}$
  2. $242$g $mol^{-1}$
  3. $224$g $mol^{-1}$
  4. $422$g $mol^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Graham's Law of Effusion, the rate of effusion is inversely proportional to the square root of molar mass. Given the conditions and pressure, the molar mass of the unknown gas calculates to 224 g/mol.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

For $S{O _2}C{l _{2\left( g \right)}} \to S{O _{2\left( g \right)}} + C{l _{2\left( g \right)}},$ Pressures of $S{O _2}C{l _2}$ at $t = 0$ and $t = 20$ minutes respectively are $700mm$ and $350mm.$ When $\log \left( {{P _0}/p} \right)$ is plotted against time ($t$), slope equals to:

  1. $1.505 \times {10^{ - 2}}{s^{ - 1}}$
  2. $1.202 \times {10^{ - 3}}{\min ^{ - 1}}$
  3. $1.505 \times {10^{ - 2}}{\min ^{ - 1}}$
  4. $0.3465\ {\min ^{ - 1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$k=\dfrac{2.303}{t}log\dfrac{P _0}{P}$ [1st order reaction]

Plot of $log\dfrac{P _0}{P}$ against $t$ will give slope $=  \dfrac{k}{2.303}$
Using given data, $k=\dfrac{2.303}{20}log \dfrac{700}{350}$
$\dfrac{k}{2.303}=\dfrac{log2}{20}=1.5\times 10^{-2}s^{-1}$
Slope $=1.5\times 10^{-2}s^{-1}$

Multiple choice biology physical resources air and winds wind energy air as a natural resource

The average amount of CO$ _2$ in the atmosphere is

  1. 100 ppm

  2. 400 ppm

  3. 1000 ppm

  4. 10 ppm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Carbon dioxide is an important gaseous component of the environment. It is fixed by the green plants by the process of photosynthesis. It is released during expiration by the process of respiration. As proclaimed by World Meteorological Organization, the average amount of carbon dioxide which can be passed globally by the atmosphere is 400 ppm. Increase in the carbon dioxide concentration may lead to global warming which is mainly caused due to human activities like deforestation.
So, the correct answer is option B.
Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

To find the concentration of $SO _2$ in the air (in parts, per

million), the data was collected for 30 localities, in a certain city

and is presented below:

Concentration of $SO _2$ (in ppm) Frequency
0.00-0.04 4
0.04-0.08 9
0.08-0.12 9
0.12-0.16 2
0.16-0.20 4
0.20-0.24 2

Find the mean concentrations of $SO _2$ in the air.

  1. $0.099$ ppm
  2. $0.09$ ppm
  3. $0.99$ ppm
  4. $0.0909$ ppm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the following table, to calculate mean:

$ci$ $f _i$  $x _i$  $f _ix _i$ 
$0.00-0.04$ $4$  $0.02$  $0.08$ 
$0.04-0.08$  $9$  $0.06$  $0.54$ 
$0.08-0.12$  $9$  $0.10$  $0.90$ 
$0.12-0.16$  $2$  $0.14$  $0.28$ 
$0.16-020$  $4$  $0.18$  $0.72$ 
$0.20-0.24$  $2$  $0.22$  $0.44$ 
$N=\Sigma f _i=30$          
 $\Sigma f _ix _i=2.96$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{2.96}{30}=0.0986667 \approx 0.099$
mean concentration of $SO _2$ in air is $0.099ppm$
Hence, option $A$ is correct.
Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

A quantity of $2.0\ g$ of a triatomic gaseous element was found to occupy a volume of $448\ ml$ at $76\ cm$ of $Hg$ and $273\ K$.

The mass of its each atom is _______________.

  1. $100\ amu$
  2. $5.53\times 10^{-23}\ g$
  3. $33.3\ g$
  4. $5.53\ amu$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given that
$P=76\ cm$ of $Hg =1\ atm$
$V=448\ ml=0.448\ L$
$T=273\ K$

Now,
$PV=nRT$

$n=\dfrac{PV}{RT}=\dfrac{1\times 0.448}{0.082\times 273}=0.02\ moles$

No. of moles $=\dfrac{mass}{molar\ mass}$

Molar mass $=\dfrac{mass}{mole}$

$3\times$ atomic mass $=\dfrac{2}{0.02}$

Atomic mass $=33.3\ g$

$\therefore$ mass of $1\ atom=\dfrac{33.3}{6.022\times 10^{23}}=5.53\times 10^{-23}\ g$
Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

Two gases X and Y have their molecular speed in ratio of $3:1$ at certain temperature. The ratio of their molecular masses $M _x:M _y$ is?

  1. $1:3$
  2. $3:1$
  3. $1:9$
  4. $9:1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Molecular speed v is proportional to 1/sqrt(M). If v_x/v_y = 3/1, then sqrt(M_y/M_x) = 3/1, so M_y/M_x = 9/1. Thus, M_x/M_y = 1/9.

Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

A gaseous mixture of three gases A, B and C has a pressure of $10$ atm. The total number of moles of all the gases is $10$. If the partial pressures of A and B are $3.0$ and $1.0$ atm, respectively, and if C has molecular mass of $2.0$, what is the mass of C, in g, present in the mixture?

  1. $6$
  2. $8$
  3. $12$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

In an experiment, the following four gases were produced. 11.2 L of which two gases at STP will weigh 14 g ?

  1. $N _2O$
  2. $NO _2$
  3. $N _2$
  4. $CO$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation
22.4 L of  a gas at STP$=$ 1 mole.
11.2 L of  a gas at STP$=$ 0.5 mole.
11.2 L of a gas at STP will weigh 14 g.
0.5 moles of a gas at STP will weigh 14 g.
1 mole of a gas at STP will weigh $\dfrac {1}{0.5} \times 14=28$ g.
The molecular weight of the gas is 28 g/mol.
Molecular weight of  $N _2O = 2 (14)+16=44$ g/mol.
Molecular weight of  $NO _2 =14+ 2 (16)=46$ g/mol.
Molecular weight of  $N _2 = 2 (14)=28$ g/mol.
Molecular weight of  $CO = 12+16=28$ g/mol.
Hence, 11.2 L of $N _2$ and $CO$ at STP will weigh 14 g.
Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

A gaseous alkane is exploded with oxygen. The volume of ${O} _{2}$ for complete combustion of alkane to $C{O} _{2}$ formed is in the ratio $7:4$. The molecular formula of alkane is:

  1. ${C} _{2}{H} _{6}$
  2. ${C} _{3}{H} _{8}$
  3. ${C} _{4}{H} _{10}$
  4. $C{H} _{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The balanced reaction is given below:


${C} _{n}{H} _{2n+2} +[n+\displaystyle\frac{n+1}{2}]{O} _{2}\rightarrow nC{O} _{2} +(n+1){H} _{2}O$

Given, 

$\displaystyle\dfrac{n+\dfrac{n+1}{2}}{n}=\dfrac{7}{4}\implies n=2$

Hence, the alkane is ${C} _{2}{H} _{6}$.

Hence, the correct option is $A$

Multiple choice chemistry air and atmosphere carbon dioxide - an oxide of carbon component of air - carbon dioxide oxides of non metal

The density of carbon dioxide is around:

  1. $2.98 {kg}/{{m}^{3}}$
  2. $1.98 {kg}/{{m}^{3}}$
  3. $5.0 {kg}/{{m}^{3}}$
  4. $10.0 {kg}/{{m}^{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Carbon dioxide is colorless. At low concentrations, the gas is odorless. At higher concentrations it has a sharp, acidic odor. At standard temperature and pressure, the density of carbon dioxide is around $1.98  {kg}/{{m}^{3}}$, about $1.67$ times that of air.

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

A pre-weighed vessel was filled with oxygen at $NTP$ and weighed. It was then evacuated, filled with $SO {2}$ at the same temperature and pressure and again weighed. The weight of oxygen is _____________.

  1. the same as that of $SO _{2}$
  2. $\dfrac{1}{2}$ that of $SO _{2}$
  3. twice that of $SO _{2}$
  4. $\dfrac{1}{4}$ that of $SO _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At constant temperature and pressure, an equal volume of gas contains an equal number of moles.


$\therefore$ Moles of $O _{2}=$ Moles of $SO _{2}$
$1\ mole\ O _{2}=32\ g$
$1\ mole\ SO _{2}=64\ g$
$\therefore wt$ of $O _2=\dfrac{1}{2}$ that of $SO _{2}$
Option $B$ is correct.

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

A certain vessel $X$ has water and nitrogen gas at a total pressure of 2 $atm$ at 300 $K$. All the contents of the vessel are transferred to another vessel $Y$ having half the capacity of the vessel $X$.The pressure of ${N} _{2}$ in this vessel was 3.8 $atm$ at 300 $K$.The vessel $Y$ is heated to 320 $K$ and the total pressure observed was 4.32 $atm$. The pressure of ${N} _{2}$ at 320 $K$ is :

[Assume that the volume occupied by the gases in the vessel is equal to the volume of the vessel.]

  1. 4.0

  2. 4.05

  3. 5.05

  4. 1.05
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using PV = nRT, the moles of N2 are constant. Initial state in X: P_total = 2 atm. Since volume of Y is half of X, P_N2 in Y at 300K = 2 * P_N2_initial. Given P_N2 in Y = 3.8 atm, P_N2_initial = 1.9 atm. At 320K, P_N2_new = P_N2_old * (T2/T1) = 3.8 * (320/300) = 4.0533 atm.