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States of Matter · Question 2

JEE Advanced 2025 Paper 1NumericalMedium📈 Topic in 17% papers+4
1

Molar volume (Vm) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with Vm as the variable. The ratio (in mol dm-3) of the coefficient of Vm2 to the coefficient of Vm for a gas having van der Waals constants a = 6.0 dm6 atm mol-2 and b = 0.060 dm3 mol-1 at 300 K and 300 atm is . Use: Universal gas constant (R) = 0.082 dm3 atm mol-1 K-1

Answer

-7.2 to -7

Solution

(P + aVm2 )(Vm - b) = RT gives Vm3 - (b + RTP )Vm2 + aPVm - abP = 0

Ratio = - (b + RTP ) aP = - bP + RTa

bP = 0.060 300 = 18 and RT = 0.082 300 = 24.6

Ratio = - 18 + 24.66.0 = -7.1 mol dm-3

Answer: −7.1

JEE Advanced 2025 Paper 1

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2

The correct statement(s) about intermolecular forces is(are)

  1. The potential energy between two point charges approaches zero more rapidly than the potential energy between a point dipole and a point charge as the distance between them approaches infinity.
  2. The average potential energy of two rotating polar molecules that are separated by a distance r has 1/r3 dependence.
  3. The dipole-induced dipole average interaction energy is independent of temperature.
  4. Nonpolar molecules attract one another even though neither has a permanent dipole moment.
JEE Advanced 2025 Paper 2
3

A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X and Y at 300 K is

  1. 2 2 : 3
  2. 2 2 : 1
  3. 1 : 2
  4. 2 : 1
JEE Advanced 2024 Paper 1

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