Polytropic process – expressing heat transfer in terms of work done For an ideal gas undergoing a polytropic process defined by p * v^n = constant, the heat absorbed (Q) during the process can be related to the work done (W) by which of the following expressions?

Difficulty: Medium

Correct Answer: [(gamma - n) / (gamma - 1)] * W

Explanation:


Introduction / Context:
This question tests your understanding of energy transfer in a general polytropic process for an ideal gas. The goal is to connect the heat transfer Q to the mechanical work W using the polytropic index n and the specific-heat ratio gamma = cp/cv.



Given Data / Assumptions:

  • Ideal gas behavior with constant specific heats (cp, cv) and gamma = cp/cv.
  • Process follows p * v^n = constant.
  • Sign convention: positive W is work done by the system; Q is heat added to the system.


Concept / Approach:
Apply the first law of thermodynamics: Q = ΔU + W. For an ideal gas, ΔU = m * cv * (T2 - T1). In a polytropic process, the work can be written in temperature form, which allows elimination of temperature change and writing Q directly as a multiple of W.



Step-by-Step Solution:
Start with W = m * R * (T2 - T1) / (1 - n).Compute ΔU = m * cv * (T2 - T1) = m * R * (T2 - T1) / (gamma - 1).Use first law: Q = ΔU + W.Factor m * R * (T2 - T1) and combine: Q = m * R * (T2 - T1) * [1/(gamma - 1) + 1/(1 - n)].Relate to W: since W = m * R * (T2 - T1) / (1 - n), then Q = [(gamma - n) / (gamma - 1)] * W.



Verification / Alternative check:
Special cases: for n = 1 (isothermal), Q = (gamma - 1)/(gamma - 1) * W = W, which matches the isothermal ideal-gas result where ΔU = 0. For n = gamma (adiabatic), Q = 0 as expected. For n < gamma, Q is positive if W is positive (expansion with heat input).



Why Other Options Are Wrong:

  • [(gamma - 1)/(gamma - n)] * W: reciprocal of the correct factor; contradicts the first-law derivation.
  • [(n - 1)/(gamma - 1)] * W: does not reduce correctly for isothermal or adiabatic limits.
  • Independent of W: false; Q is proportional to W with a factor set by n and gamma.
  • Q = 0 for any n ≠ 1: incorrect; only the adiabatic index n = gamma yields Q = 0.


Common Pitfalls:
Mixing sign conventions or using variable specific heats without noting that the closed-form factor assumes constant cp and cv.



Final Answer:
[(gamma - n) / (gamma - 1)] * W

More Questions from Thermodynamics

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion