Chemistry - Thermodynamics - 1 Question with Solution | TestHub

For a monoatomic ideal gas ($C_v=\frac{3}{2}R$, $C_p=\frac{5}{2}R$), the two paths from state A have equal temperature changes ($\Delta T_{AB}=\Delta T_{AC}=\Delta T$) since $T_B=T_C$.

Path AB (Isobaric, $P=constant$):

$q_{AB}=n{C}_p\Delta T=\frac{5}{2}R\Delta T$

Path AC (Linear through origin, $P\propto V⟹P{V}^{-1}=constant$):

$C=C_v+\frac{R}{1-x}=\frac{3}{2}R+\frac{R}{1-(-1)}=2R⟹q_{AC}=2R\Delta T$

Ratio Calculation:

$\frac{q_{AC}}{q_{AB}}\times 10=\frac{2R\Delta T}{\frac{5}{2}R\Delta T}\times 10=\frac{4}{5}\times 10=8$