Figure shows an isotherm and two isobars of two gases on a work done versus heat  supplied curve. The initial states of both gases are the same and the scales for the two  axes are same. Which straight line corresponds to nature of gas and process then select correct statement?

From first law of thermodynamics,
                                          $Q=W+\Delta U$
For an isothermal process, Q = W, slope of W – Q graph is 1, hence straight line 3  represents an isothermal process.
For an isobaric process,
                              $W=R\Delta T$
 $Q=C_p\Delta T=\left(\frac{f+2}{2}\right)R\Delta T$
Where $f$  is the number of degrees of freedom. Slope of W versus Q line is
                        $\frac{W}{Q}=\frac{2}{f+2}$
For monatomic gas $f=3$  and for diatomic gas $f=5$ . Thus slope of the straight line  representing the W versus Q is 2/5 for a monatomic gas and 2/7 for a diatomic gas.  Straight line 1 corresponds to a monatomic gas and straight line 2 corresponds to a  diatomic gas.
In an isochoric process work done is zero, the corresponding straight line coincide with  the horizontal axis.
In an adiabatic process heat is not absorbed, the corresponding straight line coincide with  the vertical axis.