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

The pressure of a gas filled in the bulb of constant volume gas thermometer at $0^0C$ and $100^0C$ are 28.6 cm and 36.6 cm of mercury respectively. The temperature of bulb at which pressure will be 35.00 cm of mercury will be:

  1. $80^0C$
  2. $70^0C$
  3. $55^0C$
  4. $40^0C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

from$T=\dfrac{p-p _0}{p _{100}-p _0}\times 100^0=\dfrac{35-28.6}{36.6-28.6}\times 100^0=80^0$

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

The pressure of a gas filled in the bulb of a constant volume gas thermometer at $0^o$C and $100^o$C are $28.6cm$ and $36.6cm$ of mercury respectively. The temperature of bulb at which pressure will be $35.0cm$ of mercury will be 

  1. $80^o$C
  2. $70^o$C
  3. $55^o$C
  4. $40^o$C
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
In a constant volume gas thermometer, the pressure of the gas varies in proportion to the temperature of the gas. The temperature varies linearly wrt to the pressure and vice versa.

 So the formula is:   $ T = \dfrac{(P - P _{0})}{(P _{100} - P _{0})}\times 100^\circ C$

     Where $T$ is the temperature at pressure $P$.
                $P _0$ is pressure at $T _0 = 0^\circ C$ 
             and $P _{100}$ is pressure at $T _{100} = 100⁰C$
                 
   $T = \dfrac{(35.0 - 28.6)}{(36.6 - 28.6)}  \times 100 ^\circ C$  
     
$  = 80^\circ C$

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

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

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

If the mass of a body is $12.1  g$ and the density is $2.2  {g}/{cc}$, its volume is

  1. $5.5 {cm}^{3}$
  2. $8 cc$
  3. $11 cc$
  4. $55 cc$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 Density = $\dfrac {mass}{volume}$ 


 so volume =$\dfrac{mass}{density}$

hence volume =$\dfrac{12.1}{2.2} =5.5cc$

hence option (A) is correct