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JEE Advanced Gaseous and Liquid States Important Questions: JEE Advanced is the toughest engineering entrance exam conducted in India. Gaseous and liquid states is an essential chapter for JEE Advanced exam. Every year, many questions appear in the exam from this chapter. To score good marks in the JEE Advanced exam, candidates need to know the states of matter of JEE Advanced questions. The page contains gaseous state JEE Advanced questions and solutions to help students understand the question’s kinds and patterns.
Candidates will get all the states of matter JEE Advanced questions of the previous year and solutions for states of matter JEE Advanced. Students can analyse their problem-solving abilities by solving these questions. They will be able to earn better scores on the entrance exam. Read the article to learn more about the JEE Advanced gaseous and liquid states important questions and download the PDFs.
– The provisional answer keys of JEE Advanced 2022 will be released on September 3, 2022. The final JEE Advanced answer key will be released on September 11, 2022.
– JEE Advanced exam 2022 was conducted on August 28, 2022.
Before getting into the specifics of JEE Advanced gaseous and liquid states important questions candidates must go through the JEE Advanced exam overview. The table below gives out a description of the JEE Advanced exam dates and other important highlights:
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Here are some of the commonly asked questions on gaseous and liquid states in JEE Advanced exam, that candidates must practice while preparing for the exam:
Question 1. An ideal gas’s diffusion coefficient is proportional to its mean free path and means speed. An ideal gas’s absolute temperature is increased four times, and its pressure is increased two times. As a result, this gas’s diffusion coefficient increases by x times. What is the value of x?
Solution: Let Di = initial value and Df = final value
Di / Df = λi √vi / λf √vf
Di / Df = Ti / Pi √Ti / M ÷ Tf / Pf √Tf / M
Di / Df = Pf .Ti √Ti ÷ Pi .Tf √Tf
Di / Df = 2Pi .Ti √Ti ÷ Pi .4Ti √4Tf
Di / Df = 2 / 8
Di / Df = ¼
Df = 4Di
Therefore, x = 4
Question 2: A closed tank has two oxygen-filled compartments, A and B. (assumed to be ideal gas). The wall separating the two sections is permanently installed and serves as an excellent heat insulator. The volume(in m3) of compartment A once the system reaches equilibrium is ________ if the previous partition is replaced with a new partition that may move and conduct heat but does not allow the gas to leak.
We know that the ideal gas equation is,
PV = nRT
⇒ n = PV / RT
If we look at Figure 1, in compartment A the pressure is 5 bar, the volume is 1 m3, and the temperature is 400 K. So we can write it as:
PA = 5
VA = 1
TA = 400
⇒ nA = 5 × 1 / 400 × R = 1 / 80 R
Likewise, in compartment B, the pressure is 1 bar, the volume is 3 m3, and the temperature is 400 K. We can write it as:
PB = 1
VB = 3
TB = 300
⇒ nB = 1 × 3 / 300 × R = 1 / 100 R
After the system attains equilibrium;
PA / TA = PB / TB
⇒ nA × R / VA = nB × R / VB
⇒ (1 / 80R) / VA × R = (1 / 100R) / VB × R
⇒ VB = ⅘ VA ………(i)
Now, the total volume of the containers.
(V) = VA + VB = 1 + 3 = 4
⇒ VA + VB = 4 ………(ii)
Now, from equations (i) and (ii), we get,
VA + ⅘ VA = 4
⇒ 9 / 5 VA = 4⇒ VA = 2.22 m3
Question 3: The equation P(V – b) = RT is satisfied by one mole of a monatomic gas. In this case, b is a constant. For the gas, the connection between interatomic potential V(r) and interatomic distance r is?
Solution: The given equation is P(V − b) = RT.
If it is compared with van der Waals’ equation, i.e., [P + a / V2] [V−b] = RT, we get a = 0.
As a result, only repulsive forces are present, and they only contribute at a very near range.
As an outcome, the potential energy rises sharply, indicating that graph (C) is true. The compressibility factor can be used to prove the repulsive force’s superiority.
P(V − b) = RT
PV = Pb + RT
PV / RT = Pb / RT +1
Z = Pb / RT +1, i.e., Z>1 (Repulsive forces).
When interatomic distances are tiny, repulsive tendencies will dominate. It also indicates that the interatomic potential is never negative, but becomes positive at short distances between the atoms.
Question 4: The volatile liquids X and Y have molar weights of 10 gmol-1 and 40 gmol-1, respectively. As illustrated in the figure, two cotton plugs, one soaked in X and the other in Y, are put at the ends of a tube with a length of L = 24 cm. At 1 atmospheric pressure and 300 degrees Fahrenheit, the tube is filled with an inert gas. At the distance of d cm from the soaked-in X plug, vapours of X and Y react to generate a product, which is first seen at a distance of d cm from the soaked-in X plug. Assume that X and Y have the same molecular diameters and that the inert gas and the two vapours behave perfectly. The value of d in cm as estimated from Graham’s law is?
Solution: The rate of diffusion is proportional to 1 / Molecular − weight
Let the distance covered by X be ′d′ and the distance covered by Y is ′24 − d′
rx / ry= d / 24 − d
⟹ d / 24 − d = √40 / 10
⟹ d = 16cm
The distance travelled (d) will be in the ratio of 2:1 since the time taken for both gases must be the same. The tube is divided into three pieces, each measuring 8 cm in length. As a result, the distance d = 16 cm (according to the 2:1 ratio of distances/velocities).
Question 5: The plot of potential energy vs internuclear distance (d) of the H2 molecule in the electronic ground state is shown below. What will be the value of the net Potential Energy Eo (as shown in the figure) in kJ mol-1 when the electron-electron and nucleus-nucleus repulsion energies are absent for d = do? When the electron and nucleus of the H atom are infinitely far apart, the potential energy of the atom is taken to be 0. 6.023 1023 mol-1 is the Avogadro constant.
Solution: P.E of 2 H-atoms
Total Energy = P.E./2
P.E = 2 T.E
E = -13.6 × z2/n2 ev/atom
= –2 × 13.6 × z2/n2 ev/atom + (–2 × 13.6 × z2/n2) ev/ atom
= –2 × 2 × 13.6 × 1 x ev/atom
= –4 × 13.6 × 1.6 × 10-19 J/atom × 6.023 × 1023 atom/mole
= –4 × 13.6 × 1.6 × 6.023 ×104 J/mole
= –5242.42 KJ/mol
Question 6. Which of the following statements about a molecule’s root mean square speed (Urms) and average translational kinetic energy (av) in gas at equilibrium is/are correct?
Solution: The following relationship exists between temperature and molecular mass and the root mean square speed (Urms) and average translational kinetic energy (av).
ϵav = √3 / 2 RT, Urms = √3RT / M
& Urms ∝ √1 / M
Therefore ϵav does not depend on its molecular mass.
Suppose we examine the above-mentioned relationship; when the temperature increases by 2 times, ϵav doubles. As a result, we can conclude that option B is inaccurate.
Therefore, the correct options are A, C, and D.
Question 7: The qualitative drawings I, II, and III below show how the surface tension of three distinct aqueous solutions of KCl, CH3OH, and CH3(CH2)11OSO3–Na+ changes with molar concentration at ambient temperature. What is the correct sketch assignment?
Solution: Impurities alter the surface tension of a liquid to some extent. Some of them, which tend to concentrate on the surface of liquids, lower the surface tension compared to the liquid’s bulk.
Surface tension is significantly reduced by compounds like detergents and soaps CH3(CH2)11OSO3–Na+. Alcohol (e.g., CH3OH, C2H5OH), on the other hand, just marginally lowers the surface tension.
The reason for this is that CH3OH has a lower dielectric constant. The surface tension is proportional to the dielectric constant. When CH3OH is added to water, the overall dielectric constant lowers, and the surface tension decreases.
Question 8: What is the ratio of the most probable, average, and root mean square speeds, respectively, if the distribution of molecular speeds of gas is as given in the figure below?
Solution: The graph depicts a symmetrical speed distribution. The most likely and average speeds should be the same. However, the average speed must be greater than the root mean square speed.
According to the molecular speed curve distribution,
Umps = √2RT / mol.wt. Uavg = √8RT / ?mol.wt. Urms = √3RT / mol.wt.
Umps : Uavg : Urms = √2 : √8 / ? : √3
= 1 : 1.128 : 1.224
Question 9: The experimental value of d is determined to be less than Graham’s law estimate. It is because of;
Solution: X collides with the inert gas with a higher frequency than Y collides with the inert gas.
The experimental value of d is found to be less than Graham’s law estimate. It is due to X colliding with the inert gas more frequently than Y colliding with the inert gas.
The molecular speed drops as the collision frequency increases, resulting in a shorter distance traversed.
Thus, D is the answer.
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Here candidates can read about the most frequently asked questions about JEE Advanced gaseous and liquid states.
Ans: Chapters from both class 11 and class 12 are included in the JEE Advanced test syllabus. As a result, Gaseous And Liquid State Important Questions JEE Advanced is essential because it covers the complete syllabus. From a scoring standpoint, these questions are critical.
Ans: The gaseous state is one of the easiest chapters to study. When the temperature varies, other quantities such as volume and pressure fluctuate as well. The properties of gases can be explained using the Kinetic theory of gases. As a result of these factors, gases are the most straightforward condition to examine.
Ans: The relationships between four parameters, namely mass, pressure, volume, and temperature, are central characteristics of gases.
Ans: Physical Chemistry by O P Tandon is a must-have for anyone studying the gaseous and liquid state for the JEE Advanced test.
Ans: The kinetic theory of gases is immensely popular among IIT JEE Advanced Physics students. The Kinetic theory of gas solutions might help candidates finish their assignments and prepare for tests swiftly.
We hope that this information on the important questions on gaseous and liquid states asked in JEE Advanced has helped you. If you have any queries, you can visit our website and write to us with your queries. We will assist you in the best way possible.
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