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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2019 · Paper I

Q1(c). Internal pressure (∂ U/∂ V)_T of a van der Waals gas

  • 10 marks
  • Answered
Question

van der Waals' equation states that (P+a V²)( V-b)=RT (a and b are constants).
Find ((∂ u)/(∂ v))_T for van der Waals' gas.

Answer in brief

((∂ Uₘ)/(∂ Vₘ))_T=a V²: positive, and set by the attraction constant a alone.

Key points

  • dU=T dS-P dV. Divide by dV at constant T: ((∂ U)/(∂ V))_T=T((∂ S)/(∂ V))_T-P
  • ((∂ S)/(∂ V))_T=((∂ P)/(∂ T))_V. Hence the thermodynamic equation of state ((∂ U)/(∂ V))_T=T((∂ P)/(∂ T))_V-P
  • P=(RT)/( V-b)-a V², so ((∂ P)/(∂ T))_V=(R)/( V-b)
  • T·(R)/( V-b)-((RT)/( V-b)-a V²), ((∂ Uₘ)/(∂ Vₘ))_T=a V²

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

The printed u and v are read as the molar energy and molar volume, Uₘ and V.

  1. Fundamental equation: dU=T dS-P dV. Divide by dV at constant T: ((∂ U)/(∂ V))_T=T((∂ S)/(∂ V))_T-P
  2. Maxwell relation (from dA=-S dT-P dV): ((∂ S)/(∂ V))_T=((∂ P)/(∂ T))_V. Hence the thermodynamic equation of state ((∂ U)/(∂ V))_T=T((∂ P)/(∂ T))_V-P
  3. van der Waals gas: P=(RT)/( V-b)-a V², so ((∂ P)/(∂ T))_V=(R)/( V-b)
  4. Substitute: T·(R)/( V-b)-((RT)/( V-b)-a V²), ((∂ Uₘ)/(∂ Vₘ))_T=a V²
  5. Meaning: this is the internal pressure. Pulling molecules apart against their attraction raises U; b (repulsion) drops out. For an ideal gas a=0 and U does not depend on volume (Joule).
  6. Integrated form: Uₘ=Uₘ^(ideal)(T)-(a)/( V), so an isothermal expansion from V₁ to V₂ needs Δ Uₘ=a((1)/( V₁)-(1)/( V₂))>0.

Application. in a Joule expansion into a vacuum (q=w=0, Δ U=0) the energy needed comes from the kinetic energy of the molecules, so a van der Waals gas cools: Δ T=-aC_(V,m)((1)/( V₁)-(1)/( V₂)). The related Joule–Thomson throttling is the basis of gas liquefaction.

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Model answer to Q1(c), 2019 Paper I: solution page 5
Solution page 5 of 40
Read the whole 2019 solution2019 question paper

All parts of Question 1