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PUCET (Punjab) Thermodynamics & Kinetic Theory — practice questions — page 2

Explore 47 free Thermodynamics & Kinetic Theory questions in PUCET (Punjab) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Thermodynamics & Kinetic Theory questions and exercises with answers

Showing questions 31–47 of 47.

If the mean free path of N₂ at 1 atm is ~80 nm, what is it at 0.01 atm (same T)?Read the answer and worked solution →The number of moles of oxygen ($O_2$) gas in a vessel of volume 1 L at STP (273.15 K, 1.013 × 10⁵ Pa) is approximately:Read the answer and worked solution →NEET · Physics · Kinetic Theory — SortRead the answer and worked solution →How free are the moleculesRead the answer and worked solution →Does the energy depend on itRead the answer and worked solution →Type the kinetic termRead the answer and worked solution →Type the kinetic numbersRead the answer and worked solution →Pop what the theory explainsRead the answer and worked solution →One word is wrongRead the answer and worked solution →Order the kinetic argumentRead the answer and worked solution →Monatomic or diatomicRead the answer and worked solution →Which one does not belongRead the answer and worked solution →Proportional or independentRead the answer and worked solution →Spot the false statementRead the answer and worked solution →Which ones belongRead the answer and worked solution →Knock out the true statementsRead the answer and worked solution →Kinetic theory bingoRead the answer and worked solution →

Worked practice questions

If the mean free path of N₂ at 1 atm is ~80 nm, what is it at 0.01 atm (same T)?

  1. ~ 8 nm
  2. ~ 80 nm
  3. ~ 800 nm
  4. ~ 8 μm (mean free path ∝ 1/P)
Show answer and solution

Correct answer: D. ~ 8 μm (mean free path ∝ 1/P)

λ ∝ T/P. Reducing P by 100× increases λ by 100×: 80 nm × 100 = 8000 nm = 8 μm. (Why vacuum coating works at low pressure — gas molecules don't collide before reaching the substrate.)

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The number of moles of oxygen ($O_2$) gas in a vessel of volume 1 L at STP (273.15 K, 1.013 × 10⁵ Pa) is approximately:

  1. ~ 0.045 mol
  2. ~ 0.045 g
  3. ~ 1.0 mol
  4. ~ 22.4 mol
Show answer and solution

Correct answer: A. ~ 0.045 mol

$n = PV/(RT) = (1.013\times10^5)(10^{-3})/(8.314\times273.15) = 101.3/2270.6 \approx 0.0446$ mol. Equivalently, 22.4 L = 1 mol at STP, so 1 L ≈ 1/22.4 = 0.0446 mol.

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NEET · Physics · Kinetic Theory — Sort

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Items
  1. Item
    Higher temperature gas
    Answer
    left
  2. Item
    Cooled gas at fixed P
    Answer
    right
  3. Item
    Lighter molecules at same T
    Answer
    left
  4. Item
    Heavier molecules at same T
    Answer
    right
  5. Item
    Hydrogen at 300 K vs Oxygen at 300 K
    Answer
    left
  6. Item
    Argon at 100 K vs Argon at 400 K
    Answer
    right
Categories
Left
Faster molecules
Right
Slower molecules
Instructions
Sort each scenario: which gas has faster average molecules?

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