Efficiency="net work done by gas"/"heat absorbed by gas" `=(|W_1|+|W_2|-|W_3|-|W_4|)/(|Q_1|)` …(i)
Process 1 On this isothermal expansion process, the constant temperature is `T_1` so work done by the gas
`W_1=nRT_1 In((V_b)/(V_a))` …(ii)
Remember that `V_b gt V_a`, so this quantity is positive, as expected. (In process 1, the gas does work by lifting something)
In isothermal process,
`Q=W`
`:.` `|Q_1|=|W_1|` ...(iii)
Process 2 On this adiabatic expansion process, the temperature and volume are related through
`TV^(gamma-1)=` constant
`:.` `T_1V_b^(gamma-1)=T_2V_c^(gamma-1)`
or `T_1/T_2=(V_c/V_b)^(gamma-1)` ...(iv)
Work done by the gas in this adiabatic process is
`W_2=(p_cV_c-p_bV_b)/(1-gamma)=(nRT_c-nRT_b)/(1-gamma)`
`=((T_2-T_1)/(1-gamma))nR=((T_1-T_2)/(gamma-1))nR` ... (v)
once again, as expected, this quantity is positive.
Process 3 In the isothermal compression process, the work done by the gas is
`W_3=nRT_(2) In(V_d)/(V_c)` ...(vi)
Because `V_d gt V_c`, the work done by the gas is negative. The work done on the gas is
`|W_3|=-nRT_(2) I n(V_d)/(V_c)=nRT_(2) In(V_c)/(V_d)` ...(vii)
Furthermore, just as in process 1,
`|Q_3|=|W_3|` ...(viii)
Process 4 In the adiabatic compression process, the calculateions are exactly the same as they were in process 2 but course with different variables. Therefore,
`T_2/T_1=(V_a/V_d)^(gamma-1)` ...(ix)
and `W_4=(nR)/(gamma-1)(T_2-T_1)` ...(x)
As expected, this quantity is negative and
`|W_4|=-W_4=(nR)/(gamma-1)(T_1-T_2)` ...(xi)
We can now calculate the efficiency.
Efficiency `=(|W_1|+|W_2|-|W_3|-|W_4|)/(|Q_1|)`
`=1+(|W_2|-|W_3|-|W_4|)/(|Q_1|)` (as `|W_1|=|Q_1|`)
But our calculations show that `|W_2|=|W_4|`.
Efficiency=`1-(|W_3|)/(|Q_1|)=1-(nRT_(2)ln(V_c//V_d))/(nRT_(1)ln(V_b//V_a))=1-(T_(2)ln(V_c//V_d))/(T_(1)ln(V_b//V_a))` ...(xii)
We have seen that
`(T_1)/(T_2)=(V_c/V_b)^(gamma-1)=(V_d/V_a)^(gamma-1)` or `V_c/V_b=V_d/V_a` or `V_c/V_d=V_b/V_a`
Substituting in Eq. (xii), we get
Efficiency `=1-T_(2)/T_(1)`
