Which one is the correct option for the two different thermodynamic processes?

Solution

Physics sir · 3 steps

Isothermal means T is fixed, so on any graph with T on the x-axis it has to be a vertical line.

1. Isothermal shape. On a P-V graph, PV = constant gives a falling hyperbola. On any graph with T on the horizontal axis, the isothermal is a vertical line, because T does not change.
2. Adiabatic shape. Start from PV^γ = constant and use PV = nRT to swap in T. Since γ > 1, P rises with T along a curve that bends upward, and V falls as T rises.

   P ∝ T^(γ/(γ-1)), V ∝ T^(-1/(γ-1))

   Graph	Isothermal drawn as	Adiabatic drawn as	Verdict
   (a) P-V	falling hyperbola, correct	vertical line, which is constant V (isochoric)	wrong
   (b) P-T	horizontal line, which is constant P (isobaric)	rising curve, correct	wrong
   (c) V-T	vertical line, correct	falling curve, correct	correct
   (d) P-T	vertical line, correct	rising curve bending upward, correct	correct

3. So. Both curves are right only in graphs (c) and (d), which is option (b).

The idea to keep. Read the axes first. Whatever stays constant in a process gives a straight line perpendicular to that axis. For an adiabatic process, rewrite PV^γ = constant in the variables on the axes.

Answer from the NTA final answer key. Solution by PrepWiser.