Chemistry · Physics

Gases and Gas Laws

252 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 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
C Correct answer
Explanation

According to the balanced reaction 2SO2 + O2 -> 2SO3, 2 moles of SO2 react with 1 mole of O2. Since we start with 2 moles of each, SO2 is completely consumed first, making it the limiting reactant. Because 3 moles of gaseous reactants form 2 moles of gaseous products, the total moles of gas decrease, leading to a decrease in the volume of the system at constant pressure.

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$

Multiple choice chemistry metals physical properties of metals and non-metals some physical properties of metals general characteristics and uses of metals

Based on the information in each choice below, which of the noble gases would have the greatest density for a $1.00\ L$ sample at room temperature and pressure?

  1. Krypton molar mass $83.8 g/mol$, atomic radius $111$ picometers
  2. Argon molar mass $39.9 g/mol$, atomic radius $94$ picometers
  3. Neon molar mass $20.2 g/mol$, atomic radius $70$ picometers
  4. Helium molar mass $4.0 g/mol$, atomic radius $50$ picometers
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ans : A Krypton molar mass $83.3\ g/ mol$, atomic radius III picometers.

Note: [ As we go top to bottom, Atomic mass increases. This is because the $e^{\ominus}$ in bigger noble gases are located away from the nucleus and held less tightly by the atom. Therefore the density of krypton is more then $He, Ne$ and $Ar$].

Multiple choice chemistry matter around us measurement of properties effect of temperature and pressure on states of matter general introduction: importance and scope of chemistry

One atmosphere is numerically equal to approximately:

  1. ${ 10 }^{ 6 }\quad dyne\quad { cm }^{ -2 }$
  2. ${ 10 }^{ 2 }\quad dyne\quad { cm }^{ -2 }$
  3. ${ 10 }^{ 4 }\quad dyne\quad { cm }^{ -2 }$
  4. ${ 10 }^{ 8 }\quad dyne\quad { cm }^{ -2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P _{dyne/cm^2}=P _{atm}\times 1.01325\times 10^6$


$P _{atm}=1$


then $P _{dyne cm^2}=1.01325\times 10^6$


Option A is Correct.

Multiple choice chemistry matter around us measurement of properties effect of temperature and pressure on states of matter general introduction: importance and scope of chemistry

1 atm equals to:

  1. 1 torr

  2. 76 torr

  3. 760 torr

  4. 100 torr
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
To convert an atmosphere measurement to a torr measurement, multiply the pressure by the conversion ratio. One atmosphere is equal to 760 torr, so use this simple formula to convert:

torr = atmospheres $\times$ 760
The pressure in torr is equal to the atmospheres multiplied by 760.

Hence, the correct option is $C$
Multiple choice chemistry matter around us measurement of properties effect of temperature and pressure on states of matter general introduction: importance and scope of chemistry

Which of the following relationships are wrong?

  1. $1 \,atm\approx 0.1 \, bar$
  2. $1 \,liter= 1 \, dm^3$
  3. $1\, J =0.239 \, cal$
  4. $1\, eV =9.11 \times 10^{-4} J$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

$1\quad atm=1.01325\quad bar\approx 1\quad bar\neq 0.1\quad bar$

$1\quad litre={ 10 }^{ -3 }{ m }^{ 3 }=1{ dm }^{ 3 }$
$1\quad cal =4.18\quad J \Rightarrow 1\quad J =0.239\quad cal\ 1\quad ev=1.6\times { 10 }^{ -19 }J \neq  9.11\times { 10 }^{ -4 }J$
Therefore, $(A). (D)$ are the correct answer.