# Volume $\;V_{S}\;$ Temperature graph of two moles of helium gas is show in figure . The ratio of heat absorbed and the work done by the gas in process 1 - 2 is

$(a)\;3\qquad(b)\;\large\frac{5}{2}\qquad(c)\;\large\frac{5}{3}\qquad(d)\;\large\frac{7}{2}$

Explanation :
Since $\;V \; \propto \;T\;$ , or $\;\large\frac{V}{T}\;$ = Constant
$\bigtriangleup Q = nC_{P} \bigtriangleup T = \large\frac{nrR}{r-1} \bigtriangleup T$
$\bigtriangleup W = P \bigtriangleup V = n R \bigtriangleup T$
$\large\frac{\bigtriangleup Q}{\bigtriangleup W} = \large\frac{r}{r-1} = \large\frac{\large\frac{3}{2}}{\large\frac{3}{2} -1}$
$= 3$