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11.Thermodynamics
medium
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
A
$3$
B
$2.5$
C
$1.67$
D
$3.5$
Solution
(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 $\frac{{\Delta Q}}{{\Delta W}} = \frac{1}{{1 – 3/5}} = \frac{5}{2}$
Standard 11
Physics
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