Find the efficiency of a cycle consisting of two isochoric and two adiabatic lines, if the volume of the ideal gas changes `n = 10` times within the cycle. The working substance is nitrogen.

Explanation

`Q'_2 = C_V (T_2 - T_3) = (C_V)/(R) V_2 (p_2 - p_3)`
`Q_1 = (C_V)/(R) V_1 (p_1 - p_4)`
Thus `eta = 1 -(V_2(p_2 - p_3))/(V_1(p_1 - p_2))`
On the other hand,
`6 p_1 V_1^gamma = p_2 V_2^gamma, p_3 V_2^gamma = p_4 V_1^gamma also V_2 = n V_1`
Thus `p_1 = p_2 n^gamma, p_4 = p_3 n^gamma`
and `eta = 1 - n^(1 - gamma)`, with `gamma = (7)/(5)` for` N_2` this is eta `= 0.602`.
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