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
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(b) $V-T$ graph is a straight line passing through origin. Hence, $V \propto T$ or $P = {\rm{constant}}$
$\therefore $ $\Delta Q = n{C_P}\Delta T$ and $\Delta U = n{C_V}\Delta T$
Also $\Delta W = \Delta Q - \Delta U = \mu \,({C_P} - {C_V})\,\Delta T$
$\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 $\frac{{\Delta Q}}{{\Delta W}} = \frac{1}{{1 - 3/5}} = \frac{5}{2}$
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