An ideal gas whose adiabatic exponent equals `gamma` goes through a cycle consisting of two isochoric and two isobaric lines. Find the efficiency of such a cycle, if the absolute temperature of the gas rises `n` times both in a isochoric heating and in the isobaric expansion.

Explanation

Since the absolute temperature of the gas rises `n` times both in the isochoric heating and in the isobaric expansion
`p_1 = np_2` and `V_2 = n V_1`. Heat taken is
`Q_1 = Q_11 + Q_12`
where `Q_11 = C_p (n - 1)T_1` and `Q_12 = C_V T_1 (1 -(1)/(n))`
Heat rejected is
`Q'_2 = Q'_21 + Q'_22` where
`Q'_21 = C_V T_1 (n -1), Q'_22 = C_p T_1 (1 -(1)/(n))`
Thus `eta = 1 -(Q'_2)/(Q_1) = 1 -(C_v (n -1) + C_p(1 - (1)/(n)))/(C_p (n -1) + C_V (1 -(1)/(n)))`
=`1 -(n -1 + gamma(1 - (1)/(n)))/(gamma (n -1) + (1 - (1)/(n))) = 1 - (1 +(gamma)/(n))/(gamma + (1)/(n)) = 1 - (n + gamma)/(1 + n gamma)`.
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