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Thermodynamics question

2022 · 27 Jul · Shift 1 · Q17
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Thermodynamics question

2022 · 27 Jul · Shift 1 · Q17

JEE MainChemistryThermodynamicsNumerical+4 / −1
The molar heat capacity for an ideal gas at constant pressure is 20.785 J K−1 mol−120.785 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}20.785 J K−1 mol−1. The change in internal energy is 5000 J5000 \mathrm{~J}5000 J upon heating it from 300 K300 \mathrm{~K}300 K to 500 K500 \mathrm{~K}500 K. The number of moles of the gas at constant volume is ‾\underline{\hspace{2cm}}​. [Nearest integer] (Given: R=8.314 J K−1 mol−1\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}R=8.314 J K−1 mol−1)
Numerical answer
View written solutionFree

Correct answer: 2

  1. Given data
  • Molar heat capacity at constant pressure: Cp=20.785 J mol−1K−1C_p = 20.785\ \text{J mol}^{-1}\text{K}^{-1}Cp​=20.785 J mol−1K−1
  • Temperature change: T1=300 K,T2=500 KT_1=300\ \text{K},\quad T_2=500\ \text{K}T1​=300 K,T2​=500 K so ΔT=500−300=200 K\Delta T = 500-300 = 200\ \text{K}ΔT=500−300=200 K
  • Change in internal energy: ΔU=5000 J\Delta U = 5000\ \text{J}ΔU=5000 J
  • Gas constant: R=8.314 J mol−1K−1R=8.314\ \text{J mol}^{-1}\text{K}^{-1}R=8.314 J mol−1K−1
  1. Find molar heat capacity at constant volume

For an ideal gas, Cp−Cv=RC_p - C_v = RCp​−Cv​=R Hence, Cv=Cp−R=20.785−8.314=12.471 J mol−1K−1C_v = C_p - R = 20.785 - 8.314 = 12.471\ \text{J mol}^{-1}\text{K}^{-1}Cv​=Cp​−R=20.785−8.314=12.471 J mol−1K−1

  1. Use the formula for change in internal energy

For an ideal gas, ΔU=nCvΔT\Delta U = n C_v \Delta TΔU=nCv​ΔT Substitute the values: 5000=n×12.471×2005000 = n \times 12.471 \times 2005000=n×12.471×200

  1. Solve for number of moles

n=500012.471×200n = \frac{5000}{12.471\times 200}n=12.471×2005000​

First compute the denominator: 12.471×200=2494.212.471\times 200 = 2494.212.471×200=2494.2

So, n=50002494.2≈2.0047n = \frac{5000}{2494.2} \approx 2.0047n=2494.25000​≈2.0047

  1. Nearest integer

n≈2n \approx 2n≈2

Therefore, the number of moles is: 2\boxed{2}2​

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