Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process 1-2 is 

 

$V-T$  graph is a straight line passing through origin. Hence,  $V\alpha T$ or  P=constant

$\therefore \Delta Q=n{C}_p\Delta T and \therefore \Delta U=n{C}_v\Delta T$

Also $\therefore \Delta W=\Delta Q-\Delta U=\mu (C_p-C_v)\Delta T$

$\therefore \frac{\Delta Q}{\Delta W}=\frac{n{C}_p\Delta T}{n(C_p-C_v)\Delta T}=\frac{C_p}{C_p-C_v}=\frac{1}{1-\frac{C_v}{C_p}}$

$\frac{C_v}{C_p}=\frac{3}{5} for helium gas. Hence \frac{\Delta Q}{\Delta W}=\frac{1}{1-3/5}=\frac{5}{2}$