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Thermodynamics question

2023 · 30 Jan · Shift 2 · Q19
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  5. /2023 · 30 Jan · Shift 2 · Q19

Thermodynamics question

2023 · 30 Jan · Shift 2 · Q19

JEE MainChemistryThermodynamicsNumerical+4 / −1
1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27∘C27^{\circ} \mathrm{C}27∘C. The work done is 3 kJ mol−13 \mathrm{~kJ} \mathrm{~mol}^{-1}3 kJ mol−1. The final temperature of the gas is ‾K\underline{\hspace{2cm}}\mathrm{K}​K(Nearest integer). Given CV=20 J mol−1 K−1\mathrm{C_V}=20 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}CV​=20 J mol−1 K−1
Numerical answer
View written solutionFree

Correct answer: 150

  1. Given data

    • Number of moles: n=1n=1n=1
    • Initial temperature: T1=27∘C=300 KT_1=27^\circ\mathrm{C}=300\,\mathrm{K}T1​=27∘C=300K
    • Adiabatic reversible expansion
    • Work done: w=3 kJ mol−1=3000 J mol−1w=3\,\mathrm{kJ\,mol^{-1}}=3000\,\mathrm{J\,mol^{-1}}w=3kJmol−1=3000Jmol−1
    • CV=20 J mol−1 K−1C_V=20\,\mathrm{J\,mol^{-1}\,K^{-1}}CV​=20Jmol−1K−1

  2. Use first law of thermodynamics

    For an adiabatic process, q=0q=0q=0

    Hence, ΔU=q−w=−w\Delta U = q-w = -wΔU=q−w=−w (taking www as work done by the gas)

    So, ΔU=−3000 J\Delta U = -3000\,\mathrm{J}ΔU=−3000J

  3. Relate internal energy change to temperature change

    For 1 mole of ideal gas, ΔU=nCV(T2−T1)\Delta U = nC_V(T_2-T_1)ΔU=nCV​(T2​−T1​)

    Substituting values, −3000=1×20×(T2−300)-3000 = 1\times 20\times (T_2-300)−3000=1×20×(T2​−300)

  4. Solve for T2T_2T2​

    T2−300=−300020=−150T_2-300 = \frac{-3000}{20} = -150T2​−300=20−3000​=−150

    Therefore, T2=300−150=150 KT_2 = 300-150 = 150\,\mathrm{K}T2​=300−150=150K

  5. Final answer 150 K\boxed{150\,\mathrm{K}}150K​

This matches the stored correct answer.

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