Two moles of a monoatomic ideal gas is taken through a cyclic process as shown in figure, process $C A$ follows $p T =$ constant .If cycle efficiency is $1 - \frac{3 x}{12 l n 2 + 15}$ ,then find the value of $x$

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Solution

$W_{C A} = (\frac{8 k}{k - 1}) R$

$= (\frac{V_{2}}{V_{2}}) \times \frac{8.314 \times 300}{2}$

$= 1.25471 k j \times 2$
$V_{C A} = 15.25 - 1.2431 \times 2$
$= 18 j$
eff $= 1 - (\frac{18}{12 ln (2) + 15})$

$= 1 - \frac{3 \times x}{1.2 l n (2) + 15}$

$3 x = 18$
$x = \frac{18}{3} = 6$

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Two moles of a monoatomic ideal gas is taken through a cyclic process as shown in figure, process CA follows pT=constant .If cycle efficiency is 1−3x12ln2+15,then find the value of x

WCA=(8kk−1)R=(V2V2)×8.314×3002=1.25471kj×2VCA=15.25−1.2431×2=18jeff=1−(1812ln(2)+15)=1−3×x1.2ln(2)+153x=18x=183=6

Two moles of a monoatomic ideal gas is taken through a cyclic process as shown in figure, process $C A$ follows $p T =$ constant .If cycle efficiency is $1 - \frac{3 x}{12 l n 2 + 15}$ ,then find the value of $x$

$W_{C A} = (\frac{8 k}{k - 1}) R$

$= (\frac{V_{2}}{V_{2}}) \times \frac{8.314 \times 300}{2}$

$= 1.25471 k j \times 2$
$V_{C A} = 15.25 - 1.2431 \times 2$
$= 18 j$
eff $= 1 - (\frac{18}{12 ln (2) + 15})$

$= 1 - \frac{3 \times x}{1.2 l n (2) + 15}$

$3 x = 18$
$x = \frac{18}{3} = 6$