MCQ
The efficiency of the cycle shown below in the figure (consisting of one isobar, one adiabat and one isotherm) is $50 \%$. The ratio $x$, between the highest and lowest temperatures attained in this cycle obeys (the working substance is an ideal gas)
  • A
    $x=e^{x-1}$
  • $x^{2}=e^{x}-1$
  • C
    $x=e^{x^{2}-1}$
  • D
    $x^{2}=e^{x^{2}-1}$

Answer

Correct option: B.
$x^{2}=e^{x}-1$
b
$(b)$ The given $p-V$ diagram can be shown as,

For curve $A B$ (isobaric),

$\frac{V_{B}-V_{A}}{T_{B}-T_{A}}$ ...........$(i)$

For curve $B C$ (adiabatic),

$T_{B}^{\top} V_{B}^{\gamma-1} =T_{C} V_{C}^{\gamma-1}$ ..........$(ii)$

$\text { Since, } V_{B} =x V_{A}$

$\text { and } V_{C} =x\left(\frac{\gamma}{\gamma-1}\right) V_{A}$

Efficiency of cycle,

$\eta=1-\frac{Q_{C A}}{Q_{A B}}$

$=1-\frac{n R T_{A} \ln \left(\frac{V_{C}}{V_{A}}\right)}{\frac{n \gamma}{\gamma-1)}\left(T_{B}-T_{A}\right)}$

$=1-\frac{\left(\frac{\gamma}{\gamma-1}\right) \ln x}{\left(\frac{\gamma}{\gamma-1}\right)(x-1)} \quad\left[\because \frac{T_{B}}{T_{A}}=x\right]$

Given, $\eta=50 \%=\frac{1}{2}$

$\Rightarrow \quad \frac{1}{2}=\frac{\ln x}{x-1} \Rightarrow x^{2}=e^{x-1}$

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