The efficiency of the cyclic process is given by then find out the value of n. pØh; izØe gS rksn

JEE MainPhysicsModerateThermodynamics

The efficiency $\eta$ of the cyclic process
$\left( \gamma = \frac{5}{3} \right)$ is given by $\frac{50n}{23}$ then
find out the value of n.
pØh; izØe $\eta\frac{50n}{23}$ gS rksn

Show answer and solution

\eta = \frac{W}{+ Q}

Temperature at A is T_0 then

P_{0}V_{0} = {nRT}_{0}

T_0gS rks

P_{0}V_{0} = {nRT}_{0}

T_{A} = T_{0},T_{B} = 2T_{0},T_{C} = 6T_{0},T_{D} = 3T_{0}

W = 2P_{0}V_{0}

+ Q = Q_{AB} + Q_{BC}

= n \cdot C_{v}dT + nC_{p}dT

= \frac{3}{2}{nRT}_{0} + \frac{5}{2}nR\left( 4T_{0} \right)

= \frac{23}{2}{nRT}_{0} = \frac{23}{2}P_{0}V_{0}

\eta = \frac{2P_{0}V_{0}}{\frac{23}{2}P_{0}V_{0}} = \frac{4}{23}

Source: Grass-G / Ph-26 Physics paper 1 15 Jan.docx

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