KVPY 2020-21 SB SX Exam Previous Papers

Question : 31 of 120

Marks: +1, -0

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 a ideal gas) :-

$x={e}^{x-1}$
$x^{2}={e}^{x-1}$
$x=e^{x^{2}-1}$
$x^{2}=e^{x^{2}-1}$

Validate

Solution:

{C_{{P}}}/{{C}_{{P}}-{R}}=\γ

{C}_{{P}}=\γ{C}_{{P}}-\γ{R}

\γ{R} \⇒(\γ-1) {C}_{{p}}

{\γ{R}}/{\γ-1}={C}_{{P}}

\η={\text "Work done "}/{\text " Heat supplied "}

\η={{nC}_{{P}} \Δ{T}-{nRT} \l{n}({{P}_{1}}/{{P}_{2}})}/{{nC}_{{P}} \Δ{T}}={1}/{2}

{nC}_{{p}} \Δ{T}=2 {nRT} l{n} {{P}_{1}}/{{P}_{2}}

({{P}_{1}}/{{P}_{2}})^{1-\γ}=({{T}_{1}}/{{T}_{2}})^{\γ}

{\γ{R}}/{\γ-1}({T}_{2}-{T}_{1})=2 {RT}_{1} \ln ({{T}_{1}}/{{T}_{2}})^{{\γ}/{1-\γ}}

{T}_{2}-{T}_{1}=2 {T}_{1} \l{n}({{T}_{2}}/{{T}_{1}})

{x}-1=\ln ({x}^{2})

{x}^{2}={e}^{x-1}

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