Question

For an ideal gas graph is shown for three processes. Process 1, 2 and 3 are respectively.

Options

• Isobaric, adiabatic, isochoric  

• Adiabatic, isobaric, isochoric

• Isochoric, adiabatic, isobaric 

• Isochoric, isobaric, adiabatic

Answer 1

Isochoric, isobaric, adiabatic

Explanation:

Isochoric processes dV = 0

W = 0 processes 1

Isobaric: W = PΔV = nRΔT

Adiabatic  |W| = `("nR"Delta"T")/(gamma-1)` 0 < γ - 1 < 1

As work is done more in the case of adiabatic processes, process 3 is adiabatic and process 2 is isobaric.