Chapter 5 · JEE Preparation Chemistry

Thermodynamics

35 practice questions · 5 marks weightage

Topics in this chapter

  • Thermodynamic Properties and First Law
  • Heat, Work, and Calorimetry
  • Enthalpy and Thermochemical Cycles
  • Entropy and Second Law
  • Gibbs Energy and Equilibrium
  • Ideal Gas Thermodynamic Processes

Cheat sheet

Cheat Sheet

Practice questions

  1. Question 1Q-B2B791

    For an isolated system, the change in internal energy (ΔU\Delta U) and change in entropy (ΔS\Delta S) respectively are:

    1. A.ΔU=0,ΔS=0\Delta U = 0, \Delta S = 0
    2. B.ΔU>0,ΔS<0\Delta U > 0, \Delta S < 0
    3. C.ΔU=0,ΔS>0\Delta U = 0, \Delta S > 0
    4. D.ΔU<0,ΔS=0\Delta U < 0, \Delta S = 0
    See the answer and solution
  2. Question 2Q-646ED5

    Which of the following conditions is always true for a spontaneous reaction at all temperatures?

    1. A.ΔH<0,ΔS<0\Delta H < 0, \Delta S < 0
    2. B.ΔH>0,ΔS>0\Delta H > 0, \Delta S > 0
    3. C.ΔH<0,ΔS>0\Delta H < 0, \Delta S > 0
    4. D.ΔH>0,ΔS<0\Delta H > 0, \Delta S < 0
    See the answer and solution
  3. Question 3Q-1A5DCD

    Calculate the enthalpy of formation of benzene (C6H6(l)\text{C}_6\text{H}_6(\text{l})) given that the heats of combustion of benzene, carbon, and hydrogen are −3267 kJ mol−1-3267\text{ kJ mol}^{-1}, −393.5 kJ mol−1-393.5\text{ kJ mol}^{-1}, and −285.8 kJ mol−1-285.8\text{ kJ mol}^{-1} respectively.

    1. A.+48.5 kJ mol−1^{-1}
    2. B.-48.5 kJ mol−1^{-1}
    3. C.+522.6 kJ mol−1^{-1}
    4. D.-1122.4 kJ mol−1^{-1}
    See the answer and solution
  4. Question 4Q-DE7FEB

    During an isothermal reversible expansion of an ideal gas against a constant external pressure, which of the following is correct?

    1. A.ΔU=0,q=−w\Delta U = 0, q = -w
    2. B.q=0,ΔU=wq = 0, \Delta U = w
    3. C.w=0,q=ΔUw = 0, q = \Delta U
    4. D.ΔSuniverse=0\Delta S_{\text{universe}} = 0
    See the answer and solution
  5. Question 5Q-DC422A

    The work done in the open vessel at 300 K300\text{ K} when 112 g112\text{ g} of iron reacts with dilute HCl\text{HCl} is: (Fe=56 g mol−1,R=0.0821 L atm K−1mol−1\text{Fe} = 56\text{ g mol}^{-1}, R = 0.0821\text{ L atm K}^{-1}\text{mol}^{-1})

    1. A.-1200 cal
    2. B.-2400 cal
    3. C.-300 cal
    4. D.-600 cal
    See the answer and solution

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