VKC ScienceChemistry for written exams Telegram

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 · 2018 · Paper I

Q4(b)(i). Why PVᵞ is constant in an adiabatic and PV in an isothermal change

  • 5 marks
  • Answered
Question

Consider the reversible expansion of a perfect gas. Provide the physical interpretation for the fact PVᵞ= constant for an adiabatic change, whereas PV= constant for an isothermal change.

Answer in brief

In an isothermal expansion heat flows in to replace the work done, so T (and U) stay constant and P∝1/V. In an adiabatic expansion no heat enters; the work is paid for from U, so T falls as V rises and P falls faster than 1/V: P∝ V⁻ᵞ with γ=Cₚ/C_V>1.

Key points

  • For a perfect gas PV=nRT. In an isothermal reversible expansion the gas is in contact with a thermostat. The work done by the gas is matched by heat drawn in (q=-w), so U and T do not change, and PV= constant (Boyle).
  • In an adiabatic expansion q=0, so dU=nC_(V,m)dT=-P dV: the work is done at the expense of internal energy, and the gas cools.
  • C_(V,m)dT/T=-RdV/V, which integrates to TV^R/C_(V,m)= constant, i.e. TVᵞ⁻¹= constant. With T∝ PV this becomes PVᵞ=constant
  • Physically, on expansion P falls for two reasons in the adiabatic case (larger volume and lower temperature) but for only one in the isothermal case. The adiabat is therefore steeper, by the factor γ: (∂ P/∂ V)_(ad)=γ(∂ P/∂ V)_T.

The full model answer

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

  1. For a perfect gas PV=nRT. In an isothermal reversible expansion the gas is in contact with a thermostat. The work done by the gas is matched by heat drawn in (q=-w), so U and T do not change, and PV= constant (Boyle).
  2. In an adiabatic expansion q=0, so dU=nC_(V,m)dT=-P dV: the work is done at the expense of internal energy, and the gas cools.
  3. Substituting P=nRT/V: C_(V,m)dT/T=-RdV/V, which integrates to TV^R/C_(V,m)= constant, i.e. TVᵞ⁻¹= constant. With T∝ PV this becomes PVᵞ=constant
  4. Physically, on expansion P falls for two reasons in the adiabatic case (larger volume and lower temperature) but for only one in the isothermal case. The adiabat is therefore steeper, by the factor γ: (∂ P/∂ V)_(ad)=γ(∂ P/∂ V)_T.
  5. γ>1 because Cₚ>C_V: for a monatomic gas γ=5/3, for a diatomic gas about 1.4. The larger C_V (more modes to absorb energy), the smaller the cooling and the closer the adiabat lies to the isotherm.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q4(b)(i), 2018 Paper I: solution page 24
Solution page 24 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 4