Two moles of an ideal gas is contained in a cylinder fitted with a frictionless movable piston, exposed to the atmosphere, at an initial temperature To. The gas is slowly heated so that its volume becomes four times the initial value. The work done by the gas is : 

6RT_o

Since, the gas is slowly heated, it remains in equilibrium (more or less) with the atmosphere, i.e. the process takes place at a constant pressure.

Now, from : PV = nRT \(\Rightarrow\) PdV = nRdT

\(\Rightarrow P \Delta V = nR \Delta T\) ... (i)

So, \(\Delta W = nR\Delta T = (2 mol) (R) (4T_o - T_o) = 6RT_o\)

[From Eq : \(\Delta V \propto \Delta T\)