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

Solution & Explanation

\textbf{Answer:} (b) $2$ \textbf{Solution:} From the common initial point on the $P$-$V$ diagram, the four processes are distinguished by their slopes: $\bullet$ Curve 1 is vertical (constant volume) $\Rightarrow$ isochoric. $\bullet$ Curve 4 is horizontal (constant pressure) $\Rightarrow$ isobaric. $\bullet$ Curves 2 and 3 are falling curves. The isothermal follows $PV=$ const (slope $\propto -P/V$), while the adiabatic follows $P{V}^{\gamma }=$ const (slope $\propto -\gamma P/V$). Since $\gamma >1$, the adiabatic curve is steeper than the isothermal. Therefore the steeper of the two falling curves, curve 2, is the \textbf{adiabatic} process, and curve 3 is isothermal. The answer is option (b).