Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Thermodynamics question

2013 · Shift 1 · Q7
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Chemistry
  4. /Thermodynamics
  5. /2013 · Shift 1 · Q7

Thermodynamics question

2013 · Shift 1 · Q7

JEE AdvancedChemistryThermodynamicsMultiple correct+3 / −0.75
Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are)
  1. A
    ΔG\Delta GΔG is positive
  2. B
    ΔSsystem\Delta S_{system}ΔSsystem​ is positive
  3. C
    ΔSsurroundings\Delta S_{surroundings}ΔSsurroundings​ = 0
  4. D
    ΔH\Delta HΔH = 0
View written solutionFree

Correct answer: B, C, D

The problem asks for the correct thermodynamic statements regarding the formation of an ideal solution by mixing benzene and naphthalene at room temperature.

Step-by-Step Analysis:

  1. Definition of an Ideal Solution: An ideal solution is one where the intermolecular forces of attraction between molecules of different components (solute-solvent) are the same as those between molecules of the same components (solute-solute and solvent-solvent). For mixing components A and B to form an ideal solution, the A-B interactions are equal in strength to the A-A and B-B interactions.

  2. Analysis of Enthalpy Change (ΔH\\\Delta HΔH): For an ideal solution, the enthalpy of mixing (ΔHmix\\\Delta H_{mix}ΔHmix​) is zero. This is because no net energy is released or absorbed during the mixing process. The energy required to break the existing interactions (benzene-benzene, naphthalene-naphthalene) is exactly balanced by the energy released when new interactions (benzene-naphthalene) are formed. ΔH=ΔHmix=0\\\Delta H = \\\Delta H_{mix} = 0ΔH=ΔHmix​=0 Therefore, statement (D) is true.

  3. Analysis of System Entropy Change (ΔSsystem\\\Delta S_{system}ΔSsystem​): Mixing is a process that increases the randomness or disorder of the system. When two pure substances are mixed, the number of possible arrangements of the molecules increases. This leads to an increase in the entropy of the system. ΔSsystem=ΔSmix>0\\\Delta S_{system} = \\\Delta S_{mix} > 0ΔSsystem​=ΔSmix​>0 For any spontaneous mixing process, the entropy of the system will increase. Therefore, statement (B) is true.

  4. Analysis of Surroundings Entropy Change (ΔSsurroundings\\\Delta S_{surroundings}ΔSsurroundings​): The change in the entropy of the surroundings is related to the heat exchanged with the system at a constant temperature T. ΔSsurroundings=−qsysT\\\Delta S_{surroundings} = \frac{-q_{sys}}{T}ΔSsurroundings​=T−qsys​​ For a process at constant pressure, the heat exchanged by the system is equal to the enthalpy change of the system, qsys=ΔHsysq_{sys} = \\\Delta H_{sys}qsys​=ΔHsys​. Since we established that for an ideal solution, DeltaHsys=0\\\\Delta H_{sys} = 0DeltaHsys​=0, then: ΔSsurroundings=−0T=0\\\Delta S_{surroundings} = \frac{-0}{T} = 0ΔSsurroundings​=T−0​=0 Therefore, statement (C) is true.

  5. Analysis of Gibbs Free Energy Change (ΔG\\\Delta GΔG): The Gibbs free energy change for a process is given by the equation: \\\Delta G = \\\Delta H - T\\\\[S_{system} For the formation of an ideal solution:

    • \\\\[H = 0
    • \\\\[S_{system} > 0
    • T (temperature in Kelvin) is positive.

    Substituting these values into the equation: ΔG=0−T(positive value)\\\Delta G = 0 - T(\\\text{positive value})ΔG=0−T(positive value) ΔG<0\\\Delta G < 0ΔG<0 A negative value of \\\\[G indicates that the process is spontaneous. The mixing of benzene and naphthalene to form a solution is a spontaneous process. Statement (A) says \\\\[G is positive, which would imply a non-spontaneous process. Therefore, statement (A) is false.

Conclusion:

Based on the analysis, the true statements are (B), (C), and (D).

PreviousNext

More from Thermodynamics

  • A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure. The succeeding operations that enable this transformation of states are Includes diagram2013 · MCQ
  • A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure. The pair of isochoric processes among the transformation of states is Includes diagram2013 · MCQ
  • For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is(are) correct? (Take Δ… Includes diagram2012 · Multiple correct
  • Using the data provided, calculate the multiple bond energy (kJ mol -1) of a C≡C bond in C2​H2​. That energy is (take the bond energy of C-H bond as 350 kJ mol -1). 2C(s)+H2​(g)→C2​H2​2C(s)→2C(g)H2​(g)→2H(g)​ΔH=225 kJ mol−1ΔH=1410 kJ mol−1ΔH=330 kJ mol−1​…2012 · MCQ
  • The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is(are) correct? Includes diagram2012 · Multiple correct
  • Match the transformations in column I with appropriate options in column II Column I (A) CO2​(s) → CO2​(g) (B) CaCO3​(s) → CaO(s) + CO2​(g) (C) 2H → H2​(g) (D) P(white, solid) → P(red, solid) Column II (p)…2011 · MCQ
  • The species which by definition has ZERO standard molar enthalpy of formation at 298 K is2010 · MCQ
  • Among the following, the intensive property is (properties are)2010 · Multiple correct