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Thermodynamics question

2024 · Shift 1 · Q8
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Thermodynamics question

2024 · Shift 1 · Q8

JEE AdvancedChemistryThermodynamicsNumerical+4 / −1
Consider the following volume-temperature (V−T)(\mathrm{V}-\mathrm{T})(V−T) diagram for the expansion of 5 moles of an ideal monoatomic gas. JEE Advanced 2024 Paper 1 Online Chemistry - Thermodynamics Question 3 English Considering only P−V\mathrm{P}-\mathrm{V}P−V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→Z\mathbf{X} \rightarrow \mathbf{Y} \rightarrow \mathbf{Z}X→Y→Z is ‾\underline{\hspace{2cm}}​. [Use the given data: Molar heat capacity of the gas for the given temperature range, CV,m=12 J K−1mol−1\mathrm{C}_{\mathrm{V}, \mathrm{m}}=12 \mathrm{~J} \mathrm{~K}^{-1}\mathrm{mol}^{-1}CV,m​=12 J K−1mol−1 and gas constant, R=8.3 J K−1 mol−1]\left.\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]R=8.3 J K−1 mol−1]
Numerical answer
View written solutionFree

Correct answer: 40600

Step-by-step Solution:

  1. Understand the Goal: The objective is to calculate the total change in enthalpy (ΔHtotal\\\Delta H_{total}ΔHtotal​) for an ideal monoatomic gas undergoing a transformation from state X\\\mathbf{X}X to state Y\\\mathbf{Y}Y and then to state Z\\\mathbf{Z}Z.

  2. Principle of State Functions: Enthalpy (HHH) is a state function. This means the change in enthalpy depends only on the initial and final states of the system, not on the path taken. Therefore, the total change in enthalpy for the process X→Y→Z\\\mathbf{X} \rightarrow \mathbf{Y} \rightarrow \mathbf{Z}X→Y→Z is simply the enthalpy change from the initial state X\\\mathbf{X}X to the final state Z\\\mathbf{Z}Z. ΔHtotal=ΔHX→Z=HZ−HX\Delta H_{total} = \Delta H_{X \rightarrow Z} = H_Z - H_XΔHtotal​=ΔHX→Z​=HZ​−HX​

  3. Formula for Enthalpy Change of an Ideal Gas: For an ideal gas, the change in enthalpy is given by the formula: ΔH=nCp,mΔT\Delta H = n C_{p, m} \Delta TΔH=nCp,m​ΔT where:

    • nnn is the number of moles of the gas.
    • Cp,mC_{p, m}Cp,m​ is the molar heat capacity at constant pressure.
    • ΔT\Delta TΔT is the change in temperature (Tfinal−TinitialT_{final} - T_{initial}Tfinal​−Tinitial​).
  4. Extract Data from the Problem:

    • Number of moles, n=5n = 5n=5 mol.
    • Molar heat capacity at constant volume, CV,m=12 J K−1mol−1C_{V, m} = 12 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}CV,m​=12 J K−1mol−1.
    • Gas constant, R=8.3 J K−1mol−1R = 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}R=8.3 J K−1mol−1.
    • From the V-T diagram:
      • Initial temperature (at state X), TX=200 KT_X = 200 \mathrm{~K}TX​=200 K.
      • Final temperature (at state Z), TZ=600 KT_Z = 600 \mathrm{~K}TZ​=600 K.
  5. Calculate Molar Heat Capacity at Constant Pressure (Cp,mC_{p, m}Cp,m​): Using Mayer's relation for an ideal gas, Cp,m=CV,m+RC_{p, m} = C_{V, m} + RCp,m​=CV,m​+R. Cp,m=12 J K−1mol−1+8.3 J K−1mol−1=20.3 J K−1mol−1C_{p, m} = 12 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1} + 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1} = 20.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}Cp,m​=12 J K−1mol−1+8.3 J K−1mol−1=20.3 J K−1mol−1

  6. Calculate the Total Change in Temperature (ΔT\Delta TΔT): The overall process starts at state X\\\mathbf{X}X and ends at state Z\\\mathbf{Z}Z. ΔT=Tfinal−Tinitial=TZ−TX=600 K−200 K=400 K\Delta T = T_{final} - T_{initial} = T_Z - T_X = 600 \mathrm{~K} - 200 \mathrm{~K} = 400 \mathrm{~K}ΔT=Tfinal​−Tinitial​=TZ​−TX​=600 K−200 K=400 K

  7. Calculate the Total Change in Enthalpy (ΔHtotal\Delta H_{total}ΔHtotal​): Now, substitute the values of nnn, Cp,mC_{p, m}Cp,m​, and ΔT\Delta TΔT into the enthalpy change formula. ΔHtotal=nCp,mΔT\Delta H_{total} = n C_{p, m} \Delta TΔHtotal​=nCp,m​ΔT ΔHtotal=(5 mol)×(20.3 J K−1mol−1)×(400 K)\Delta H_{total} = (5 \mathrm{~mol}) \times (20.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}) \times (400 \mathrm{~K})ΔHtotal​=(5 mol)×(20.3 J K−1mol−1)×(400 K) ΔHtotal=5×20.3×400 J\Delta H_{total} = 5 \times 20.3 \times 400 \mathrm{~J}ΔHtotal​=5×20.3×400 J ΔHtotal=5×8120 J\Delta H_{total} = 5 \times 8120 \mathrm{~J}ΔHtotal​=5×8120 J ΔHtotal=40600 J\Delta H_{total} = 40600 \mathrm{~J}ΔHtotal​=40600 J

Conclusion:

The total change in enthalpy for the transformation is 40600 J.

Comparison with Stored Answer:

The calculated value is 40600 J, while the stored correct answer is 8120 J. The value 8120 J corresponds to the molar change in enthalpy (ΔHm=Cp,mΔT=20.3×400=8120 J/mol\Delta H_m = C_{p,m} \Delta T = 20.3 \times 400 = 8120 \mathrm{~J/mol}ΔHm​=Cp,m​ΔT=20.3×400=8120 J/mol), which would be the total enthalpy change if there were only 1 mole of gas. Since the question explicitly states there are 5 moles, the total enthalpy change should be 5 times the molar enthalpy change. Thus, the stored answer appears to be incorrect based on the provided problem statement.

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