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 pressure in fluids and atmospheric pressure common consequences of the atmospheric pressure atmospheric pressure and its consequences important points about atmospheric pressure

A $2-m3$ weather balloon is loosely filled with helium at $1\ atm (76\ cm\ Hg)$ and at $27^{\circ}C$. At an elevation of $20,000\ ft$, the atmospheric pressure is down to $38\ cm\ Hg$ and the helium has expanded, being under no constraint from the confining bag. If the temperature at this elevation is $-48^{\circ}C$, the gas volume (in $m3)$ is

  1. $3$
  2. $4.28$
  3. $2$
  4. $2.5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the ideal gas law in combined form, (P_1 * V_1) / T_1 = (P_2 * V_2) / T_2. Given P_1 = 76 cm Hg, V_1 = 2 m^3, T_1 = 27 + 273 = 300 K, P_2 = 38 cm Hg, and T_2 = -48 + 273 = 225 K. Solving for V_2 gives V_2 = (P_1 / P_2) * (T_2 / T_1) * V_1 = (76 / 38) * (225 / 300) * 2 = 2 * 0.75 * 2 = 3 m^3.

Multiple choice physics pressure in fluids and atmospheric pressure common consequences of the atmospheric pressure atmospheric pressure and its consequences important points about atmospheric pressure

76 cm of mercury column exerts a pressure equal to that exerted by:

  1. 7920 m of air column

  2. 105 m of air column

  3. 1 cm of air column

  4. 1.29 m of air column
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

1 cm of mercury column exerts pressure=105 m of air column
76cm of mercury column exerts pressure =105 x 76 cm = 7980 m of air column

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A gas is enclosed in a rectangular vessel. $20 \times 10^23$ molecules of the gas strike a well of the vessel normally per second, with a velocity of 250m/s and rebound with the same speed in the opposite direction. What is the force exerted by the gas on the wall if the mass of each molecule is $5 \times 10^-23$ g ? 

  1. 50 N

  2. 40 N

  3. 75 N

  4. 25 N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force = change in momentum per unit time. Delta p = m * v - (m * -v) = 2mv. Force = N * (2mv) / t. Given N/t = 20 * 10^23 molecules/s, m = 5 * 10^-26 kg (converting 5 * 10^-23 g), v = 250 m/s. F = 20 * 10^23 * 2 * 5 * 10^-26 * 250 = 20 * 10^23 * 10^-25 * 250 = 20 * 0.1 * 250 = 50 N.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

KE per unit volume is E. The pressure exerted by the gas is given by:

  1. $\displaystyle \frac {E}{3}$
  2. $\displaystyle \frac {2E}{3}$
  3. $\displaystyle \frac {3E}{2}$
  4. $\displaystyle \frac {E}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pressure exerted by gas is given by $P=\dfrac{1}{3}\rho v^2$

where $v$ is the velocity of gas particles, $\rho $ is the density of gas.
Kinetic energy per unit volume$=E=\dfrac{\dfrac{1}{2}mv^2}{V}=\dfrac{1}{2}\rho v^2$
Thus $P=\dfrac{2}{3}(\dfrac{1}{2}\rho v^2)=\dfrac{2E}{3}$

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A cylinder at a certain ternperature has a gas at a pressure ot $50cm$ of Hg.Then it is divided into three equal parts so that the gas in the central part is completely transferred to either equally. Find the pressure of the gas in each portion.

  1. $10cm$
  2. $20cm$
  3. $75cm$
  4. $100cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The length of vacuum above mercury column in a barometer is $10cm/cc$ of air from outside where the pressure is $76cm$ of Hg is passed into the barometer tube. The area of cross section of the tube is $1{ cm }^{ 2 }$.The height of the mercury column then will be

  1. $71cm$
  2. $72cm$
  3. $73cm$
  4. $74cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A thermally insulated cylinder is divided into two equal halves by a thermally insulated wall. One half contains He at 600$\mathrm { K }$ and the other contains $\mathrm { H } _ { 2 }$ at 800$\mathrm { K }$ , pressure being same in two parts equal to $P _ { 0 }$ . Now, the wall is removed and two gases mix. The resulting pressure is

  1. $P _ { 0 }$
  2. 2$P _ { 0 }$
  3. $\frac { 28 P _ { 0 } } { 54 }$
  4. $\frac { 28 P _ { 0 } } { 27 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Dalton's law and energy conservation: P_final = (n_He*R*T_He + n_H2*R*T_H2) / (V_total). Since P0 = nRT/V, n = P0*V/RT. P_final = (P0*V/R*600 * R*600 + P0*V/R*800 * R*800) / (2V) is incorrect. Correct: P_final = (n1+n2)RT_mix / 2V. Using internal energy: (n1Cv1T1 + n2Cv2T2) = (n1Cv1 + n2Cv2)T_mix. For monoatomic (He) Cv=3/2R, diatomic (H2) Cv=5/2R. Result is 28P0/27.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

Two identical cylinders contain Helium at 2.5atmosphere and Argon at 1 atmosphere respectively. If both gases are transferred in one ofthe cylinders, what is the new pressure? 

  1. $3.5$ atmosphere
  2. $1.5$ atmosphere
  3. $1.75$ atmosphere
  4. $1$ atmosphere.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the cylinders are identical and we are transferring gases into one, we use Dalton's Law of partial pressures. P_total = P1 + P2 = 2.5 + 1 = 3.5 atm.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A box is divided into two equal compartments by a thin partition and they are filled with gases $  P  $and $  Q  $ respectively. The two compartments have a pressure of 250 torr each. The pressure after removing the partition will be equal to

  1. $125 torr$
  2. $2.5 torr$
  3. $250 torr$
  4. $500 torr$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the partition is removed, the total volume doubles (V + V = 2V). Since the pressure in each was 250 torr, the total moles are proportional to 250*V + 250*V = 500*V. New pressure = 500V / 2V = 250 torr.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

Two vessels of volume 100 c.c and 150 c.c contains gases at pressure of 1 atm and 2 atm. When they are joined the common pressure is 

  1. 2 atm

  2. 1.5 atm

  3. 1.6 atm

  4. 1 atm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Boyle's law and conservation of mass for ideal gases at constant temperature, the final common pressure P is given by (P1V1 + P2V2) / (V1 + V2). Substituting the values: (1 * 100 + 2 * 150) / (100 + 150) = (100 + 300) / 250 = 400 / 250 = 1.6 atm.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

$28\, gm$ of $N _2$ gas is contained in a flack at a pressure of $10\, atm$ and at a temperature of $57^0$. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to $27^0 C$. The quantity of $N _2$ gas that leaked out is :-

  1. $\dfrac{11}{20}\, gm$
  2. $\dfrac{20}{11}\, gm$
  3. $\dfrac{5}{63}\, gm$
  4. $\dfrac{63}{5}\, gm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the ideal gas law PV = nRT, we calculate the initial moles of N2 (n1 = P1V1 / RT1) and final moles (n2 = P2V2 / RT2). Since the volume is constant, the ratio of moles is (P2/P1) * (T1/T2). Calculating the difference in moles and converting to mass gives 63/5 grams.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The pressure and temperature of an ideal gas in a closed vessel are $720$ kpa and $40^oC$ respectively. If - th of the gas is released from the vessel and the temperature of the remaining gas is raised to $353^oC$, the final pressure of the gas is 

  1. $ 1440$ kPa
  2. $1080$ kPa
  3. $720$ kPa
  4. $ 540$ kPa
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial: P1 = 720 kPa, T1 = 313 K. After releasing 1/3 of gas, remaining gas is 2/3 of original: n2 = (2/3)n1. Temperature rises to T2 = 353 + 273 = 626 K. Using P2V = n2RT2: P2 = (n2/n1)(T2/T1)P1 = (2/3)(626/313)(720) = (2/3)(2)(720) = 960 kPa. Wait, this doesn't match. Let me recalculate: if 1/3 is released, remaining is 2/3. Temperature ratio: (353+273)/(40+273) = 626/313 = 2. So P2 = (2/3) × 2 × 720 = 960 kPa. This doesn't match option B (1080 kPa). Perhaps 1/3 remaining means released 2/3? Then P2 = (1/3) × 2 × 720 = 480 kPa. Still no match. Checking if -th means 1/4 released (3/4 remaining): P2 = (3/4) × 2 × 720 = 1080 kPa. Yes! This matches.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A container holds $ 10^{26} molecules / m^3 $ each of mass $ 3 \times 10^{-27} $ Kg. Assume that 1/6 of the ,molecule move with  velocity 2000 m/s directly towards one wall of the container while the remaining 5/6 of the molecules move either away from the wall or in perpendicular direction, and all collision of the molecules with the wall or in perpendicular direction, and all collision of the molecules with the wall are elastic.

  1. Number of molecules hitting $ 1 m^2 $ of the wall every second is $ 3 .33 \times 10^{28} $
  2. Number of molecules hitting $ 1 m^2 $ of the wall every second is $ 2 \times 10^{29} $
  3. Pressure exerted on the wall by molecules is $ 24 \times 10^5 Pa. $
  4. Pressure exerted on the wall by moleculaes is $ 4 \times 10^5 Pa, $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice zoology human influences on the environment carbon cycle ozone layer depletion and causes depletion of ozone layer

At present, the concentration of atmospheric $CO 2$ is ________.

  1. $100$ ppm
  2. $240$ ppm
  3. $380$ ppm
  4. $520$ ppm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
CO2 is a major pollutant of air and a green house gas that contributes to global warming. Concentration of CO2 in the atmosphere is expressed in terms of parts per million or ppm. According to latest research by forecasting systems, CO2 concentration was found to be about 380 ppm which in recent years has increased upto 400 ppm.
So, the correct answer is '380 ppm'.
Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Which of the following is the density of nitrogen at STP?

  1. $0.33g/L$
  2. $0.65g/L$
  3. $0.80g/L$
  4. $1.25g/L$
  5. $1.60g/L$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At STP, one mole of any gas occupies 22.4litres of volume. Nitrogen exists in molecular form as $N _2$. 

So, $Denisty= \dfrac{\text Molecular \ mass}{\text Volume}$

Here a Molecular mass of one mole of $N _2$. is 14(2)=28 grams and Volume is 22.4 litres.
So $Denisty= \dfrac{28}{22.4}g/L = 1.25g/L$.
Option D is correct