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 substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Maximum density of $H _2O$is at the temperature

  1. $32^oF$
  2. $39.2^oF$
  3. $42^oF$
  4. $4^oF$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Maximum density of water is at $4^0C$


Here, the value in degree celsius is converting to degree Fahrenheit.

$T(°F) = (T(°C) × \dfrac 95 )+ 32$

or

$T(°F) = (T(°C) × 1.8) + 32$

We have,

$T(^0C)=4^0C$

Then,

$T(°F) = (4 × 1.8) + 32=39.2^oF$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A $10L$ container is filled with a gas to a pressure of $2atm$ at $0^0C.$ At what temperature will the pressure inside the container be $2.50atm?$   

  1. $68^0C$
  2. $50^0C$
  3. $40^0C$
  4. $45^0C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since at both times, container is same . which means that volume remains constant. By applying ideal gas equation.

$PV=nRT$

or $\dfrac{P _{1}}{T _{1}}=\dfrac{P _{2}}{T _{2}}$

Substituting as per question,
$P _{1}=2 atm$ 
$p _{2}=2.5atm$
$T _{1}=0^\circ or \ 273k \ use \ S.I.unit $
$T _{2}=?$

$\dfrac{2}{273}=\dfrac{2.5}{T _{2}}$ or $T _{2}=\dfrac{2.5\times 273}{2}$

$=341.25K$           $(341.25-273=68.25)$
$=68.25^\circ C$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

The density of mercury is $13600 kg m^{-3} $.Its  value in CGS system will be:

  1. $13.6 g cm^{-3}$
  2. $1360 g cm^{-3}$
  3. $136 g cm^{-3}$
  4. $1.36 g cm^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The $C.G.S.$ unit of kg is Gm$.$ and that of meter is cm$.$ When we convert $13600\,kg{m^{ - 3}}$ then $13600 \times 1000\left( {100 \times 100 \times 100} \right) = 13.6\,gc{m^{ - 3}}$

Hence,
option $(A)$ is correct answer.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A vessel contains a mixture consisting of ${m} _{1}=7kg$ of nitrogen $\left( { M } _{ 1 }=28 \right) $ and ${m} _{2}=11g$ of carbon dioixide $\left( { M } _{ 2 }=44 \right) $ at temeprature $T=300K$ and pressure ${ P } _{ 0 }=1\quad atm$. The density of the mixture is:

  1. $1.446g$ per litres
  2. $2.567g$ per litre
  3. $3.752g$ per litre
  4. $4.572g$ per litre
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let V is the volume of the vessel.

Now, let $p _{1}$ and $p _{2}$ be the partial pressure, then using gas law: 

$p _{1}V = \dfrac{m _1}{M _1}RT\\$

$p _{2}V = \dfrac{m _2}{M _2}RT,\ p _{0}  = p _{1} +  p _{2}\\$

$p _{0} = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{V}\\$

$V = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{p _{0}}\\$

$\because \rho _{mix}=\dfrac{(m _{1} + m _{2})}{V}\\$

$rho _{mix}=\dfrac {(m _1 + m _2)M _1 M _2} {(m _1M _2 + m _2M _1)} \times \dfrac{p _0}{RT}\\$

Substituting values,

$\rho _{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$

$= 1.446 \ per \ litre$

Option A is correct.

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

When you combust $100.0\ g$ of propane at $500.K$ and $1.00\ atm$ in a closed container, you expected to collect $279\ L$ of carbon dioxide. Instead, when you collect the gas, it measures $651\ L$ in total.
Why have you collected more than your predicted, theoretical yield?

  1. The theoretical yield was calculated incorrectly.

  2. The graduated cylinder used to collect the gas was read incorrectly.

  3. You did not account for the surrounding volume of air.

  4. You did not account for the volume of water vapor that was produced.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\text{The theoretical yield was calculaterd incorrectly.}$

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

Equal masses of hydrogen gas and oxygen gas are placed in a closed container at a pressure of $3.4 atm$. The contribution of hydrogen gas to the total pressure is:

  1. $1.7 atm$
  2. $0.2 atm$
  3. $3.2 atm$
  4. $3.02 atm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let mass of hydrogen and oxygen be $100g$

$n 1=$ no of moles of hydrogen=$\cfrac {100}{2}=50$
$n _2=$ no of moles of oxygen=\cfrac {100}{32}=3.125$
Contribution of hydrogen to the total pressure means, mole fraction of hydrogen present in the mixture (partial pressure of $H_2$)
$X_4=\cfrac {n{H_2}}{n_{H_2}+n_{O_2}}=\cfrac {50}{50+3.125}=0.94$
Contribution of hydroegn to the total pressure= $0.94\times P=0.94 \times 3.4=3.2$ atm

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

A closed vessel contains equal number of oxygen and hydrogen molecules at a total pressure of 740 mm. If oxygen is removed from the system, the pressure -

  1. Becomes half of 740 mm.

  2. Remains unchanged

  3. Becomes 1/9th of 740 mm.

  4. Becomes double 740 mm.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Dalton's Law of Partial Pressures, the total pressure is the sum of the partial pressures of the gases. If the number of molecules is equal, each gas contributes half the total pressure (370 mm each). Removing oxygen leaves only hydrogen, so the pressure becomes 370 mm, which is half of 740 mm.

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

A closed vessel contains equal number of nitrogen and oxygen molecules at a pressure of P mm. If nitrogen is removed from the system, then the pressure will be?

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

$\dfrac{p _1}{n _1}=\dfrac{p _2}{n _2}$ (As molecules of $N _2=$ molecules $O _2$)
($\therefore$ Moles $N _2=$moles $O _2$)
As $N _2$ molecules removed$=$ moles become half
$\dfrac{p _1}{1}=\dfrac{p _2}{1/2}$
$p=\dfrac{1}{2}p _1$
$\therefore$ Pressure reduced to half
$\therefore$ Answer is option C.

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

Consider the following reaction, taking place in a container fitted with a movable piston.
$2SO _{2}(g) + O _{2}(g) \rightarrow 2SO _{3}(g)$
Suppose we place two moles each of $SO _{2}$ and $O _{2}$ in the reaction vessel at $25^{\circ}C$, and adjust the volume to give a total pressure of $1.0\ atm$. The reaction is ignited by a spark, and goes to completion. The temperature is returned to $25^{\circ}$.
Which of the following best describes this system after reaction is complete?

  1. $SO _{2}$ is limiting. The volume of the system remains the same.

  2. Neither reactant is limiting. The volume of the system decreases.
  3. $SO _{2}$ is limiting. The volume of the system decreases.
  4. $O _{2}$ is limiting. The volume of the system remains the same.
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

Two closed vessel $A$ and $B$ of equal volume of $8.21L$ are connected by a narrow tube of negligible volume with open valve. The left hand side container id found to contain $3\ mole \, CO _2$ and $2\ mole$ of $He$ at $400K$. What is the partial pressure of $He$ in vessel $B$ at $500K$?

  1. 2.4 atm

  2. 8 atm

  3. 12 atm

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

partial pressure of $He$ in vessel $B$, 

$PV = nRT$
$P _{He} = \dfrac{nRT}{V} = \dfrac{2\times 8.314\times 400}{8.21} =8atm$

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

If the density of a certain gas at $30^oC$ and $768 \ torr$ is $1.35 \ kg/{m}^{3}$, the density at STP would be:

  1. $1.48 \ kg/{m}^{3}$
  2. $1.58 \ kg/{m}^{3}$
  3. $1.25 \ kg/{m}^{3}$
  4. $1.4 \ kg/{m}^{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
From Ideal gas law,

$PM = dRT$

$P _1 = 768\ torr$
$T _1 = 30^0\ C = 303\ K$
$d _1 = 1.35 kg/ m^3$

At STP, 
$T _2 = 273\ K$
$P _2 = 760\ torr$

$\dfrac{P _1}{P _2} = \dfrac{d _1T _1}{d _2T _2}$

$\dfrac{768}{760} = \dfrac{1.35\times 303}{d _2\times 273}$

$d _2 = 1.48 kg/m^3$

Hence, option A is correct.
Multiple choice evs - i the earth and its living world introduction to the spheres of the earth origin of universe introduction to natural resources

Atmospheric content of $CO _2$ is

  1. $0.0036\%$
  2. $0.036\%$
  3. $0.36\%$
  4. $3.6\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Carbon dioxide is a colourless gas with a density of about 60% higher than that of dry air. Carbon dioxide consists of a carbon atom covalently double bonded to two oxygen atoms. It occurs naturally in Earth's atmosphere as a trace gas. The current concentration is about 0.04% (410 ppm) by volume, having risen from pre-industrial levels of 280 ppm. Carbon dioxide is odourless at normally encountered concentrations. However, at high concentrations, it has a sharp and acidic odour.

So the correct answer is '0.036%'.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The gaseous mixture consists of $16\quad $ of helium and $16\quad $ of oxygen. The ratio $\cfrac { { C } _{ p } }{ { C } _{ v } } $ of the mixture is :-

  1. 1.59

  2. 1.62

  3. 1.4

  4. 1.54
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Helium is monatomic (gamma = 1.67) and Oxygen is diatomic (gamma = 1.4). Using the formula for the mixture ratio of specific heats, the result is approximately 1.62.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A gaseous mixture consists of $16\ g$ of helium and $16\ g$ of oxygen, then the ratio $\dfrac { { C } _{ p } }{ { C } _{ v } } $of the mixture is

  1. $1.4$
  2. $1.54$
  3. $1.59$
  4. $1.62$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hellium is monoatomic.

Oxygen is diatomic.
Degree of freedom of He, ${ f } _{ 1 }=$3
Degree of freedom of ${ O } _{ 2 }$, ${ f } _{ 2 }=$5
$\therefore \cfrac { { C } _{ p } }{ { C } _{ v } }$ of mixture.
$\cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { [{ N } _{ 1 }(2+{ f } _{ 1 })+{ N } _{ 2 }(2+{ f } _{ 2 })] }{ { N } _{ 1 }{ f } _{ 1 }+{ N } _{ 2 }{ f } _{ 2 } }$
Putting,${ N } _{ 1 }={ N } _{ 2 }=16\quad gm,\quad$ and $\ { f } _{ 1 }=3,\quad { f } _{ 2 }=5$
we get,$ \cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { 3 }{ 2 } \ =1.62$