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An ideal gas undergoes four different processes from the same initial state as shown in the figure. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is

Asked in NEET 2022 · Reading and Translating P-V Diagrams

Figure: Reading and Translating P-V Diagrams
Answer: (2) 2

Step-by-step solution

Idea: on a P-V diagram each process has a signature slope at a given state.

A vertical line is constant volume (isochoric); a horizontal line is constant pressure (isobaric).

So curve 1 is isochoric and curve 4 is isobaric, leaving 2 and 3 for the isotherm and the adiabat.

For an isotherm PV= constant gives ((dP)/(dV))_T=-P/V.

For an adiabat PV^γ= constant gives ((dP)/(dV))_Q=-γP/V, and γ>1.

The adiabat therefore falls more steeply from the common starting point, which is curve 2.

Why the other options are wrong

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