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Q.
For an ideal gas graph is shown for three processes.
Process 1, 2 and 3 are respectively. Work done (magnitude)
Thermodynamics
Solution:
Isochoric proceess $ d V=0$
$W=0$ proceess $1$
Isobaric : $ W=P \Delta V=n R \Delta T$
Adiabatic $\left|W_{\mid}\right|=\frac{n R \Delta T}{\gamma-1} 0 <\gamma -1 < 1$
As work done in case of adiabatic process is more so process 3 is adiabatic and process 2 is isobaric.