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September 16, 201139 Insightful Publications

**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.

**Latest Updates:**– The provisional answer keys of JEE Advanced 2022 will be released on

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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:

Exam Events | Dates |
---|---|

Registration Start Date | To Be Announced |

Registration End Date | To Be Announced |

Fee Payment Last Date | To Be Announced |

Exam Date of JEE Advanced 2023 | To Be Announced |

Provisional Answer Keys Release Date | To Be Announced |

Final Answer Keys Release Date | To Be Announced |

Result Date of JEE Advanced 2023 | To Be Announced |

Architecture Aptitude Test 2023 Registration Date | To Be Announced |

Seat Allocation by JoSAA 2023 | To Be Announced |

Architecture Aptitude Test Date 2023 | To Be Announced |

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.**

**Solution: **

We know that the ideal gas equation is,

PV = nRT

⇒ n = PV / RT

Given:

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′

Now,

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?**

**When the temperature of the Urms is increased four times, it doubles.****When the temperature of av is increased four times, it doubles.****The molecular mass of av at a certain temperature has no bearing on it.****Urms are inversely proportional to their molecular mass squared.**

**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, CH _{3}OH, and CH_{3}(CH2)_{11}OSO_{3}–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 CH_{3}(CH_{2})_{11}OSO_{3}–Na+. Alcohol (e.g., CH_{3}OH, C_{2}H_{5}OH), on the other hand, just marginally lowers the surface tension.

The reason for this is that CH_{3}OH has a lower dielectric constant. The surface tension is proportional to the dielectric constant. When CH_{3}OH 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;**

**In comparison to Y, X has a larger mean free path.****In comparison to X, Y has a larger mean free path.****The collision frequency of Y with the inert gas is higher than that of X with the inert gas.****X collides with the inert gas with a higher frequency than Y collides with the inert gas.**

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.

**Related Links:**

JEE Advanced Important Dates | JEE Advanced Admit Card |

JEE Advanced Eligibility Criteria | JEE Advanced Exam Pattern |

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