States of Matter JEE Main Questions 2025: Important Practice Questions with PYQs

States of Matter JEE Main Questions 2025 - As JEE Main 2025 is approaching, candidates should fasten their preparation and be ready with the topics of JEE Main 2025 Chemistry by now. JEE Main Chemistry 2025 is one of the easiest yet tactical subjects that helps fetch good score in JEE Mains with a proper preparation strategy. Approximately 2 questions are asked from States of Matter every year in the JEE Mains entrance exam which is equivalent to 6.6% weightage. Therefore, candidates are advised to prepare well for the chapter by practising the JEE Main 2025 States of Matter Questions. Some of the important topics that hold the highest weightage from States of Matter JEE Main are Ideal Gas Equation and Real Gases, Kinetic Theory of Gases, Boyle’s, Charles’s, Avogadro’s Laws, etc. In this article, we have discussed the States of Matter JEE Main Questions 2025 where candidates can practice the Important Questions that are extracted from Previous years questions.

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JEE Main 2025 States of Matter Important Questions

1. If Z is the compressibility factor, van der Waals’ equation at low pressure can be written as :

(1) Z=1-Pb/RT

(2) Z=1+Pb/RT

(3) Z = 1+RT/Pb

(4) Z = 1-a/VmRT

Solution:

(P+a/Vm2 )(Vm-b) = RT [Van der waals equation of state]

Vm-b ≈ Vm

So the equation becomes (P+a/Vm2 )Vm= RT

⇒PVm+a/Vm = RT

Divide all terms by RT

PVm/RT+a/VmRT = RT/RT

PVm/RT = 1- a/VmRT

Z = 1- a/VmRT [Z = PVm/RT]

Hence option (4) is the answer.

Also Read: JEE Main 2025 Preparation and Study Time Table for 60 Days (2 Months)

2. Which intermolecular force is most responsible for allowing xenon gas to liquefy?

(1) Instantaneous dipole induced dipole

(2) Ion dipole

(3) Ionic

(4) Dipole-dipole

Solution:

For the liquefaction of xenon, instantaneous dipole-induced dipole forces are responsible.

Hence option (1) is the answer.

3. The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is : (Atomic masses: He = 4 u, O = 16 u)

(1) 1200 K

(2) 600 K

(3) 300 K

(4) 2400 K

Solution:

Given Atomic masses : He = 4 u, O = 16 u

(Vrms) O2 = (Vrms) He

√(3RT1/M1) = √(3RT2/M2)

T1 /M1 = T2/M2

T1/32 = 300/4

T1 = 300×32/4

= 2400 K

Hence option (4) is the answer.

4. The compressibility factor for a real gas at high pressure is :

(1) 1-Pb/RT

(2) 1+ RT/Pb

(3) 1

(4) 1+Pb/RT

Solution:

(P+a/V2)(V-b) = RT [Real gas equation]

a/V2 can be neglected at high pressure.

PV-Pb = RT

PV/RT = (RT/RT) + (Pb/RT)

PV/RT = 1 + (Pb/RT) …(1)

Z = PV/RT …(2)

Equating (1) and (2)

Z = 1 + (Pb/RT)

Hence option (4) is the answer.

5. The relationship among most probable velocity, average velocity and root mean square velocity is respectively :

(1) √2 : √(8/π) : √3

(2) √2 :√3 : √(8/π )

(3) √3 : √(8/π): √2

(4) √(8/π) : √3: √2

Solution:

Vmpv = √(2RT/M)

Vav = √(8RT/πM)

Vrms = √(3RT/M)

Vmpv: Vav: Vrms. = √(2RT/M) : √(8RT/πM) : √(3RT/M)

= .√2 : √(8/π) : √3

Hence option (1) is the answer.

6. The value of gas constant R is

(1) 0.082 L atm

(2) 0.987 cal mol-1 K-1

(3) 8.3 J mol-1 K-1

(4) 83 erg mol-1K-1

Solution:

R = 8.3 J mol-1 K-1

Hence option (3) is the answer.

7. By how much does the temperature of a gas increase when the root mean The square velocity of the gas molecules in a container of fixed volume is increased from 5×104 cm/s to 10×104 cm/s.

(1) Four

(2) three

(3) Two

(4) Six

Solution:

Vrms ∝ √T

V1/V2 = √(T1/T2) = 5×104/10×104

squaring, we get

T1/T2 = 25/100 = ¼

T2 = 4T1

Hence option (1) is the answer.

8. The kinetic theory of gases proves

(1) Only Boyle’s law

(2) Only Charle’s law

(3) Only Avogadro’s law

(4) all of these

Solution:

One of the postulates of the kinetic theory of gases is average kinetic energy proportional to T.

This theory proves all the above-given laws.

Hence option (4) is the answer.

9. Which one of the following is the wrong assumption of the kinetic theory of gases?

(1) All the molecules move in a straight line between collision and with the same velocity.

(2) Molecules are separated by great distances compared to their sizes.

(3) Pressure is the result of elastic collision of molecules with the container’s wall.

(4) Momentum and energy always remain conserved.

Solution:

The molecules are always in random motion and obey Newton’s law of motion. They have velocities in all directions ranging from zero to infinity.

Hence option (1) is the answer.

10. ‘a’ and ‘b’ are Vander Waal’s constants for gases. Chlorine is more easily liquified than ethane because:

(1) a for Cl2 < a for C2H6 but b for Cl2 > b for C2H6

(2) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

(3) a and b for Cl2 > a and b for C2H6

(4) a and b for Cl2 < a and b for C2H6

Solution:

The greater the ‘a’ value, the more easily the gas is liquified, the lower the ‘b’ value, the more easily the gas is liquified.

Hence option (2) is the answer.

States of Matter JEE Main 2025 Important Topics

States of Matter JEE Main Questions 2025 PDF (Previous Years Questions)

Practice the important previous years' States of Matter JEE Main Questions from the following PDF provided below.

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