An ideal gas goes through a cycle consisting of alternate isothermal and adiabatic curves (Fig. 2.2). The isothermal processes proceed at the temperatures `T_1, T_2` and `T_3`. Find the efficiency of such a cycle, if in each isothermal expansion the gas volume increases in the same alphaortion.
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Explanation

Given `V_2 = n V_1, V_4 = n V_3`
`Q_1` = Heat taken at the upper temperature
=`RT_1 1n n + RT_2 1n n = R(T_1 + T_2) 1n n`
Now `T_1 V_2^(gamma - 1) = T_2 V_3^(gamma - 1)` or `V_3 = ((T_1)/(T_2))^((1)/(gamma - 1)) V_2`
Similarly `V_5 = ((T_2)/(T_3))^((1)/(gamma - 1)) V_4, V_6 = ((T_1)/(T_3))^((1)/(gamma - 1)) V_1`
Thus `Q_2` = heat ejected at the lower temperature `= -RT_3 1n (V_6)/(V_5)`
=`-RT_3 1n((T_1)/(T_2))^((1)/(gamma -1)) (V_1)/(V_4) = -RT_3 1n ((T_1)/(T_2))^((1)/(gamma -1)) (V_2)/(n^2 V_3)`
=`-RT_3 1n ((T_1)/(T_2))^((1)/(gamma - 1)) (1)/(n^2) ((T_1)/(T_2))^(-(1)/(gamma - 1)) = 2 R T_3 1n n`
Thus `eta = 1 - (2 T_3)/(T_1 + T_2)`.
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