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 physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

A thin tube sealed at both ends, is $100\ cm$ long. It lies horizontally, the middle $0.1\ m$ containing mercury and the two ends containing air at standard atmospheric pressure. If the tube is turned to a vertical position, by what amount will the mercury be displaced ?

  1. $1.84\ cm$
  2. $8.45\ cm$
  3. $2.95\ cm$
  4. $5\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the tube is vertical, the pressure difference between the two air columns must balance the weight of the mercury column. Using Boyle's Law for both air columns, we find the displacement of the mercury.

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

8 gm $O _{2}$,14 gm $N _{2}$ and 22 gm $CO _{2}$ is mixed in a container of 10 litre capacity at $27^oC$.The pressure exerted by the mixture in terms of atmospheric pressure will be-

  1. 1

  2. 3

  3. 9

  4. 18
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

First find the moles of each gas: moles of O2 = 8/32 = 0.25, moles of N2 = 14/28 = 0.5, moles of CO2 = 22/44 = 0.5. Total moles = 0.25 + 0.5 + 0.5 = 1.25 moles. Using the ideal gas law P * V = n * R * T, with V = 10 L, T = 300 K, and R = 0.0821, we find P = (1.25 * 0.0821 * 300) / 10 approximately equal to 3.07 atm, which rounds to 3 atm.

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

An air tight container having a lid with negligible mass and a area of $8 c m^{2}$ is partially evacuated . If a 40N force is require to pull a lid off the container and the atmospheric pressure is $ 1.0 \times 10^{5} \mathrm{Ps}$ ,the presure in the container before it is opened must be

  1. 0.6atm

  2. 0.5atm

  3. 0.4atn

  4. 0.2atm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The pressure inside the container must be equal to the pressure developed by force applied and the atmospheric force would oppose the pull,

So,

$P _{external}=P _{inside}+P _{atmosphere}$

$=\dfrac{48}{0.008}=P _{inside}+1\times 10^5$

$P _{inside}=0.6\times 10^5-1\times 10^5$

$P _{inside}=-0.4\times 10^5\,Ps=0.4\,atm$

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

A gas is collected over the water at $25 ^ { \circ } \mathrm { C }$ . The total pressureof moist gas was 735$\mathrm { mm }$ of mercury. If the aqueous vapourpressure at $25 ^ { \circ } \mathrm { C }$ is 23.8$\mathrm { mm }$ . Then the pressure of dry gas is

  1. $760$ $\mathrm { mm }$
  2. $758.8$ $\mathrm { mm }$
  3. $710.8$ $\mathrm { mm }$
  4. $711.2$ $\mathrm { mm }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The pressure of dry gas is calculated by subtracting the aqueous vapor pressure from the total pressure of the moist gas. Pressure_dry = Total_pressure - Vapor_pressure = 735 mm - 23.8 mm = 711.2 mm.

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

20$\mathrm { cm }$ containing mercury and two equal ends containing air at standardatmospheric prossure. If the tube is now turned to a vertical position, by what amount will the mercury bo displaced?
(Given : cross-section of the tube can be assumed to be uniform):

  1. 2.95$\mathrm { cm }$
  2. 5.18$\mathrm { cm }$
  3. 8.65$\mathrm { cm }$
  4. 0.0$\mathrm { cm }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

A small bulb is filled with air at ${ 41 }^{ 0 }c$ and sealed. When it heated slowly in an oil bath upto ${ 198 }^{ 0 }c$, it is found to be broken. find the pressure at which the bulb is broken.

  1. 2.5 atm

  2. 1.5 atm

  3. 3 atm

  4. 1.3 atm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a sealed bulb at constant volume, Gay-Lussac's Law applies: P1 / T1 = P2 / T2. Initial temperature T1 = 41 + 273 = 314 K, initial pressure P1 = 1 atm. Final temperature T2 = 198 + 273 = 471 K. P2 = (1 * 471) / 314 = 1.5 atm.

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

One gram mole of oxygen is enclosed in a vessel at a temperature of $27$ and at one atmospheric pressure. The vessel is thermally insulated and is moved with a constant speed $u _0$. Calculate $u _0$ if the rise in temperature is $1K$ when the vessel is suddenly stopped?

  1. $35.6 m/s$
  2. $45 m/s$
  3. $90 m/s$
  4. $60 m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

since internal energy depends only on the temperature ,

as the temperature changes,
the change in internal energy is
$\Delta U= \dfrac{nfR \Delta T}{2}$

vessel contains a gas of mass M,
change in kinetic energy $= \dfrac{nMv^2}{2}$

$\dfrac{nfR \Delta T}{2}$= $\dfrac{nMv^2}{2}$

$\dfrac{2}{\dfrac{c _p}{c _v} - 1}$

$f = \dfrac{2}{\dfrac{2}{5}}$ = 5

putting value

$v = 45 m/s$

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

If 5 litres of kerosene has a mass of 5 kg, then what is the density of kerosene?

  1. 500 kg/$m^3$
  2. 1000 kg/$m^3$
  3. 100 kg/$m^3$
  4. 50 kg/$m^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Density of kerosene $= \displaystyle \frac{mass}{volume}$
$\displaystyle = \frac{5  kg}{5  litres}= \frac{5kg}{5 \times 10^{-3}m^3} = 1000 kg  m^{-3}$

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

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.