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JEE Main Classification of Elements and Periodicity in Properties — practice questions

Explore 13 free Classification of Elements and Periodicity in Properties questions in JEE Main Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Classification of Elements and Periodicity in Properties questions and exercises with answers

Showing questions 1–13 of 13.

A heat engine operates with 20% efficiency between two reservoirs. If T_h = 500 K, find T_c (for maximum/Carnot):Read the answer and worked solution →An ideal gas undergoes isothermal expansion at 300 K from V = 1 to V = 2. Heat absorbed (n = 1):Read the answer and worked solution →A diatomic ideal gas at 27°C is compressed adiabatically to ½ its volume. Find final temperature (γ = 7/5):Read the answer and worked solution →Glucose in open-chain form has a:Read the answer and worked solution →Which atom has the larger radius?Read the answer and worked solution →Which has the higher first ionization energy?Read the answer and worked solution →Match the functional group to its nameRead the answer and worked solution →Organic reactions — true or false?Read the answer and worked solution →Spot the odd type of isomerismRead the answer and worked solution →Match the thermodynamic quantity to its symbolRead the answer and worked solution →Order the calorimetry steps for heat of neutralisationRead the answer and worked solution →Complete the Gibbs free energy formulaRead the answer and worked solution →Entropy — true or false?Read the answer and worked solution →

Worked practice questions

A heat engine operates with 20% efficiency between two reservoirs. If T_h = 500 K, find T_c (for maximum/Carnot):

  1. 450 K
  2. 300 K
  3. 400 K
  4. 100 K
Show answer and solution

Correct answer: C. 400 K

We have η = 1 - T_c/T_h → 0.20. That is η = 1 - T_c/500 → T_c/500. That is η = 0.80 → T_c. So η = 400 K.

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An ideal gas undergoes isothermal expansion at 300 K from V = 1 to V = 2. Heat absorbed (n = 1):

  1. R × 300 × ln(1/2)
  2. R × 300
  3. R × 300 × ln 2
  4. R × 600
Show answer and solution

Correct answer: C. R × 300 × ln 2

In isothermal expansion at T, Q = W = nRT ln(Vf/Vi) = 1 × R × 300 × ln(2/1) = 300R ln 2 ≈ 208R J.

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A diatomic ideal gas at 27°C is compressed adiabatically to ½ its volume. Find final temperature (γ = 7/5):

  1. 600 K
  2. 395 K
  3. 240 K
  4. 420 K
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Correct answer: B. 395 K

T₁V₁^(γ-1) = T₂V₂^(γ-1). T₁ = 300 K, V₂ = V₁/2. T₂ = T₁(V₁/V₂)^(γ-1) = 300 × 2^0.4 ≈ 300 × 1.32 ≈ 396 K. Adiabatic compression heats the gas.

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More Chemistry topics

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