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 real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

If pressure of ${CO} _{2}$ (real gas) in a container is given by $P=\cfrac { RT }{ 2V-b } -\cfrac { a }{ 4{ b }^{ 2 } } $, then mass of the gas in container is:

  1. $11g$
  2. $22g$
  3. $33g$
  4. $44g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Van Der waal's equation for $n$ mole of real gas 


$\bigg( P +\dfrac{n^2 a}{V^2}\bigg)(V- nb)=nRT\implies P=\dfrac{nRT}{V-nb}-\dfrac{n^2a}{V^2}$

Given that Pressure of $CO _2$ gas in a contaner is given by:
$P= \dfrac{RT}{2V-b}-\dfrac{a}{4b^2}$

Compairing it with the standard Van der waal's equation we get :
$n=\dfrac12$

Therefore, Number of moles in a container , $n=\dfrac12$
Molar mass of $CO _2= 44\ gm$
Mass of gas in the container, $m= \dfrac12\times 44 =22 gm$


Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

An experiment requires a gas with $\gamma = 1.50$. This can be achieved by mixing together monatomic and rigid diatomic ideal gases. The ratio of moles of the monatomic to diatomic gas in the mixture is

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

One mole of an ideal monoatomic gas is is C$ _{v}$ = $\dfrac{3}{2}$R and C$ _{p}$ = $\dfrac{5}{2}$R


i.e $\gamma$ = 1.66 for monoatomic gas

For One mole of an ideal dioatomic gas,
$\gamma$ = 1.4 for air which is pre dominantly a  diatomic gas
If we take 1 mole monoatomic and 1 mole of diatomic gas in a mixture then we get the following result;

$\gamma$ = $\dfrac{n1\gamma + n2\gamma}{n1 + n2}$ 

Now since we have taken the no. of moles of monoatomic as well as diatomic as 1, therefore
$\gamma$ = $\dfrac{y1 + y2}{2}$ where $\gamma$1 and $\gamma$2 are the values of $\dfrac{C _p}{C _v}$ for individual gases.

Substuting the values of C$ _p$ and C$ _v$ i.e $\gamma$1 = 1.6 and $\gamma$2 = 1.4 we get
$\gamma$ = 1.53 which is approximately equal to 1.50 which is given.
Hence by taking 1 mole og monoatomic and 1 mole of diatomic mixture we got $\gamma$ as 1.50
Hence the ratio of moles of monoatomic to diatomic gas in the mixture is 1:1

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

How many degrees of freedom the gas molecules have if under STP the gas density $\rho = 1.3 kg/m^3$ and the velocity of sound propagation in it is $330 ms^{-1}$?

  1. $3$
  2. $5$
  3. $7$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using v = sqrt(gamma * P / rho) and P = (rho/M)RT, we find gamma. Since v = 330 m/s, rho = 1.3 kg/m^3, and P = 1.013e5 Pa, gamma = v^2 * rho / P = 1.71. However, for standard gases, gamma = 1 + 2/f. With gamma = 1.4 (diatomic), f = 5. The provided data yields gamma approx 1.4, corresponding to f = 5.

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

The value of $\gamma$ for gas X is 1.66, then x is :

  1. Ne
  2. O$ _3$
  3. N$ _2$
  4. H$ _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that value of gamma is 1.66 i.e $\dfrac{5}{3}$ which implies that it is a monoatomic gas, and Neon (Ne) is the only monoatomic gas among the given options.

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

A vessel contains a non-linear triatomic gas. If $50$% of gas dissociate into individual atom, then find new value of degree of freedom by ignoring the vibrational mode and any further dissociation 

  1. $2.15$
  2. $3.75$
  3. $5.25$
  4. $6.35$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let's assume we have $1$ mole of triatomic gas

$\therefore 3Na$ is present
So, $0.5$ moles= $1.5 Na$ atoms
$1$ part of $0.5$ moles remains untouched
Degree of dissociation= $0.5 \times 6=3$
Degree of freedom for $0.5Na= 1.5 \times 0.5=0.75$
Total=$3+0.75=3.75$

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

$2$ grams of mono atomic gas occupies a volume of $2$ litres at a pressure of $8.3 \times 10^5$ Pa and $127^0C$. Find the molecular weight of the gas.

  1. $2$ grams/mole
  2. $16$ grams/mole
  3. $4$ grams/mole
  4. $32$ grams/mole
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using PV = (m/M)RT, where m=2g, P=8.3e5 Pa, V=2e-3 m^3, T=400K, R=8.314. M = mRT/PV = (2 * 8.314 * 400) / (8.3e5 * 2e-3) = 6651.2 / 1660 = 4.006 g/mol.

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

A vessel contains a non-linear triatomic gas. If $50$% of gas dissociate into individual atom, then find new value of degree of freedom by ignoring the vibrational mode and any further dissociation.

  1. 2.15

  2. 3.75

  3. 5.25

  4. 6.35
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Non-linear triatomic gas has f=6. If 50% dissociates into 3 atoms, the new mixture has 0.5 moles of triatomic (f=6) and 0.5 * 3 = 1.5 moles of monatomic (f=3). Average f = (0.5*6 + 1.5*3) / (0.5 + 1.5) = (3 + 4.5) / 2 = 7.5 / 2 = 3.75.

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

A mixture Of $n _ { 2 }$ moles of mono atomic gas and $n _ { 2 }$ moles of diatomic gas has $\frac { C _ { p } } { C _ { V } } = y = 1.5$

  1. $n _ { 1 } = n _ { 2 }$
  2. $2 n _ { 1 } = n _ { 2 }$
  3. $n _ { 1 } = 2 n _ { 2 }$
  4. $2 n _ { 1 } = 3 n _ { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} As\, { y _{ mix } }=\dfrac { { { C _{ { p _{ mix } } } } } }{ { { C _{ { v _{ mix } } } } } } where\, { C _{ { p _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ and\, \, { C _{ { v _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { v _{ 1 } } } }+{ n _{ 2 } }{ C _{ { v _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ So,\, \, { y _{ mix } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ Given\, ,\, for\, monoatomic\, { C _{ p } },\dfrac { 5 }{ 2 } R\, and\, { C _{ { v _{ 1 } } } }=\dfrac { 3 }{ 2 } R \ For\, diatomic\, { C _{ { p _{ 2 } } } }=\dfrac { { 7R } }{ 2 } \, and\, { C _{ { v _{ 2 } } } }=\dfrac { 5 }{ 2 } R \ { y _{ mix } }=\dfrac { { { n _{ 1 } }\times \dfrac { 5 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 7 }{ 2 } R } }{ { { n _{ 1 } }\times \dfrac { 3 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 5 }{ 2 } R } } =\dfrac { { 5{ n _{ 1 } }+7{ n _{ 2 } } } }{ { 3{ n _{ 1 } }+5{ n _{ 2 } } } } =\dfrac { 3 }{ 2 }  \ 10{ n _{ 1 } }+14{ n _{ 2 } }=9{ n _{ 1 } }+15{ n _{ 2 } } \ { n _{ 1 } }={ n _{ 2 } } \ Hence, \ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

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

A vessel of volume $0.3 \ { { m }^{ 3 } }$ contains Helium at $20.0$. The average kinetic energy per molecule for the gas is:

  1. $6.07\times { 10 }^{ -21 }J$
  2. $7.3\times { 10 }^{ 3 }J$
  3. $14.6\times { 10 }^{ 3 }J$
  4. $12.14\times { 10 }^{ -21 }J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given temperature of gas $=20°C$
                                            $=293K$
$\Rightarrow$ Average translational kinetic energy $=\dfrac{3}{2}KT$
                                                                   $=\dfrac{3}{2}\times1.38\times10^{-23}\times293$
                                                                   $=6.07\times10^{-21}J$
Hence, the answer is $6.07\times10^{-21}J.$
Multiple choice chemistry nitrogen and sulfur sulfur and its oxides acid deposition natural resources- air, water and land

$1000L$ of air at STP was dissolved in water and required $2.5\times { 10 }^{ -5 }$ moles of $KMn{ O } _{ 4 }$ for complete reaction of ${SO} _{2}$ at pollutants. Thus, ${SO} _{2}$ content in air is :

  1. $1.4ppm$
  2. $14ppm$
  3. $2.8ppm$
  4. $6.25ppm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reaction: 5SO2 + 2MnO4- + 2H2O -> 5SO4(2-) + 2Mn(2+) + 4H+. 2 moles of MnO4- react with 5 moles of SO2. 2.5*10^-5 moles of KMnO4 react with (5/2)*2.5*10^-5 = 6.25*10^-5 moles of SO2. Volume of air = 1000 L. Moles of SO2 per L = 6.25*10^-8. At STP, 1 mole = 22.4 L. Concentration in ppm = (Volume SO2 / Volume air) * 10^6 = (6.25*10^-5 * 22.4 / 1000) * 10^6 = 1.4 ppm.

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The number of oxygen molecules in a cylinder of volume $1 \mathrm { m } ^ { 3 }$ at a temperature of $27 ^ { \circ } C$ and pressure $13.8 Pa$ is
 (Boltzmaan's constant $k = 1.38 \times 10 ^ { - 23 } \mathrm { JK } ^ { - 1 }$)

  1. $6.23 \times 10 ^ { 26 }$
  2. $0.33 \times 10 ^ { 28 }$
  3. $3.3 \times 10 ^ { 21 }$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice bio-chemistry absorption by roots - the processes involved diffusion transport across membrane means of transport

Rate of diffusion of $CO _2$ is $20$ times higher than $O _2$, as it depends upon?

  1. Solubility of gases

  2. Partial pressure

  3. Thickness of membrane

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Gas exchange is the process that occurs between oxygen and carbon dioxide. Oxygen is passed from the lungs to the bloodstream and carbon dioxide is eliminated from the bloodstream to the lungs. Exchange of Gas takes place in lungs between the alveoli and capillaries which are tiny blood vessels, placed at the walls of alveoli. The rate of diffusion depends on the thickness of the biological membrane which forms the boundary between the external environment and organisms.

Gas exchange takes place by simple diffusion based on a concentration/pressure gradient. The rate of diffusion depends not only on the solubility of gases but also on the thickness of the membranes involved in gas exchange.
So the correct answer is 'All of these'.

Multiple choice

What is the maximum allowable concentration of carbon monoxide in a mine?

  1. 50 parts per million (ppm).

  2. 100 ppm.

  3. 150 ppm.

  4. 200 ppm.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The maximum allowable concentration of carbon monoxide in a mine is 50 parts per million (ppm).

Multiple choice

According to MSHA regulations, what is the maximum permissible concentration of methane in an underground mine?

  1. 1%

  2. 2%

  3. 5%

  4. 10%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

MSHA regulations specify that the maximum permissible concentration of methane in an underground mine is 1%. Exceeding this limit poses a significant risk of explosion.