Question

Find the efficiency of the thermodynamic cycle shown in figure for an ideal diatomic gas.​

Answer 1

Answer:

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Answer 2

$\Large\bf {\orange {\underline { \underline {Solutíon}} : - }}$

Let n be the number of moles of the gas and the temperature be $\sf T_{0}$in the state A. Now, work done during the cycle

$\sf w = \frac{1}{2} \times ( 2V_{0} - V_{0})( 2P_{0} - P_{0}) = \frac{1}{2} P_{ 0 } V_{0}$

For the heat $\sf (∆Q_{1})$ given during the process A → B, we have

$\sf ∆Q_{1} = ∆W_{AB} + ∆U_{AB}$

$\sf ∆W_{AB} = area \: under \: the \: straight \: line \: AB$

$\sf = \frac{1}{2} ( P_{0} + 2P_{0})( 2V_{0} - V_{0} ) = \frac{3P_{0} V_{0} }{2}$

Applying equation of state for the gas in the state A & B.

$\sf \frac{ P_{0} V_{0} }{ T_{0} } = \frac{( 2P_{0})( 2V_{0}) }{ T_{B} } \Longrightarrow T_{B} = 4T_{0}$

$\sf \therefore U_{AB} = nC_{v}∆T = n \bigg( \frac{5R}{2} \bigg)( 4T_{0} - T_{0})$

$\sf = \frac{ 15nRT_{0} }{2} = \frac{ 15P_{0} V_{0}}{2}$

$\sf \therefore \: ∆Q_{1} = \frac{3}{2} P_{0} V_{0} + \frac{1}{5} P_{0} V_{0} = 9P_{0} V_{0}$

Obviously, the processes B → C and ⠀⠀⠀ C → A involved the abstraction of heat from the gas .

$\boxed{ \sf Efficiency = \frac{work \: done \: per \: cycle}{total \: heat \: supplied \: per \: cycle} }$

i.e., $\sf η = \frac{ \frac{1}{2}P_{0} V_{0} }{9P_{0} V_{0} } = \frac{1}{18}$