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

Which units of pressure are needed if you are going to use 0.0821 as your ideal gas constant?

  1. Atmospheric

  2. Torr

  3. Psi

  4. mmHg

  5. Pascals
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Gas constant = $0.0821 $  $L atm K^{−1} mol^{−1}$


atm ( Atmospheric unit is used )

Hence the correct option is A.

Multiple choice

Methane content of biogas is

  1. 24.6%

  2. 55.8%

  3. 8.6%

  4. 4.7%

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

Biogas is a mixture of gases, primarily methane (50-70%), carbon dioxide, and small amounts of other gases. 55.8% falls within the standard range.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

Air is filled at $60^o$C in a vessel of open mouth. The vessel is heated at temperature T so that $\dfrac{1}{4}$ th part of air escapes. Assuming volume of container remaining constant, find value of T. 

  1. $80^oC$
  2. $444^oC$
  3. $333^oC$
  4. $171^oC$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using PV = nRT, since P and V are constant, n1*T1 = n2*T2. If 1/4 escapes, n2 = 3/4 n1. So n1 * (60 + 273) = (3/4)n1 * (T + 273). 333 = 0.75 * (T + 273). 444 = T + 273, so T = 171 C.

Multiple choice chemistry matter in our surroundings diffusion in different states of matter properties of solids, liquid, and gas particle theory of matter

Consider three identical flasks with different gases:


Flask A: CO at $760$ torr and $273$ K
Flask B: $N _2$ at $250$ torr and $273$ K
Flask C: $H _2$ at $100$ torr and $273$ K

In which flask will the molecules have the greatest average kinetic energy per mole?

  1. A

  2. B

  3. C

  4. Same in all
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Avg kinetic energy$=\dfrac{3}{2}nRT$


$\therefore$ For $1$ mole:-$\dfrac{3}{2}RT$

As the temperature in all of these is same.

$\therefore$ Avg kinetic energy of all these are equal.

Hence, option $D$ is correct.

Multiple choice zoology respiratory system of human mechanism of respiration respiratory cycle breathing and exchange of gas

Fill in the blanks:


Component              Inspired air             Expired air
Oxygen                            a                         16.4%
Nitrogen                        79%                        b

  1. a = 5.6%, b = 21.6%

  2. a = 20.96%, b = 79.6%

  3. a = 28.8, b = 98%

  4. a = 1%, b = 2%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Component       Inspired air     Expired air

Oxygen                  20.96%            16.4%
Carbon dioxide       0.04%              4%
Nitrogen                   79%               79.6%

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

The number of air molecules in a $(5m\times5m\times4m)$ room at standard temperature and pressure is of the order of

  1. $6\times10^{23}$
  2. $3\times10^{24}$
  3. $3\times10^{27}$
  4. $6\times10^{30}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume V = 5*5*4 = 100 m^3. At STP, 1 mole occupies 22.4 liters (0.0224 m^3). Number of moles n = 100 / 0.0224 approx 4464 moles. Number of molecules = n * Avogadro's number = 4464 * 6e23 approx 2.68e27, which is of the order of 10^27.

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

The ratio of number of collisions per second at the walls of containers by $He$ and $O _2$ gas molecules kept at same volume and temperature, is (assume normal incidence on walls) ?

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

The rate of collisions per unit area is proportional to (n * v_avg), where n is number density and v_avg is average speed. Since n = N/V and V, T are same, n is same. v_avg is proportional to 1/sqrt(M). Ratio = v_He / v_O2 = sqrt(M_O2 / M_He) = sqrt(32 / 4) = sqrt(8) = 2*sqrt(2).

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

1 mole of $SO _2$ occupies a volume of $350 ml$ at $300K$ and $50 atm $ pressure. Calculate the compressibility factor of the gas.

  1. $0.888$
  2. $0.711$
  3. $0.520$
  4. $0.987$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $P=50 \ atm , \ \ V=350 ml =0.350 \ \ \ litre, \ \ n=1 \ \ mole $ and $ T=300 K$

Now, Compressibility factor $Z= \dfrac{PV}{nRT}$
$\therefore  \ Z= \dfrac{50 \times 0.350}{1 \times 0.082 \times 300}= 0.711$

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

If 2g of helium is enclosed in a vessel at NTP, how much heat should be added to it to double the pressure ? (Specific heat of helium = 3 J/gm K)

  1. 1638 J

  2. 1019 J

  3. 1568 J

  4. 836 J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

No. of moles, $n=\dfrac{m}{M}=\dfrac{2}{4}=0.5 mol$

The specific heat, $C _V=3J/g.mol K$
$C _V=12 J/mol. K$
At constant volume,
$\dfrac{T _2}{T _1}=\dfrac{P _1}{P _2}$
$T _2=2T _1$
$\Delta T=2T _1-T _1=T _1=273K$
The heat required, $\Delta Q=nC _V \Delta T$
$\Delta Q=0.5\times 12\times 273$
$\Delta Q=1638 J$
The correct option is A.

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

The diameter of oxygen molecules is $2.94 \times 10^{-10}m $. The Van der Waals gas constant in m$^3$/mol will be

  1. $3.2$
  2. $32$
  3. $32\times 10^{-6}$
  4. $32 \times 10^{-3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle b = 4N \times \frac{4}{3} \pi \frac{d^3}{8}$ (standard definition of boyles constant)

$\displaystyle = \frac{4 \times 6.02 \times 10^{23} \times 3.14 \times 2.94^3 \times 10^{-30}}{3 \times 8}$

$\displaystyle = 32 \times 10^{-6}$

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

The size of container B is double that of A and gas in B is at double the temperature and pressure than that in A. The ratio of molecules in the two containers will then be -

  1. $\frac{N _B}{N _A} = \frac{1}{1}$
  2. $\frac{N _B}{N _A} = \frac{2}{1}$
  3. $\frac{N _B}{N _A} = \frac{4}{1}$
  4. $\frac{N _B}{N _A} = \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the ideal gas law PV = NkT, the number of molecules N = PV/(kT). Container B has double the volume (VB = 2VA), double the pressure (PB = 2PA), and double the temperature (TB = 2TA) compared to container A. Substituting these into the ratio gives NB/NA = (PB VB TA) / (PA VA TB) = (2 * 2 * 1) / (1 * 1 * 2) = 4/2 = 2/1.

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