One mole of a diatomic ideal gas `(gamma = 1.4)` is taken through a cyclic process starting from point A. The process `A to B` is an adiabatic compression. `B to C` is isobaric expansion, `D to A` an adiabatic expansion and is isochoric.
The volume ratio are `V_A //V_B - 16` and `V_C //V_B - 2` and the temperature at `T_A = 300k` is
Find the temperature of the gas at the point (in K).

Answer

`61.4`

Explanation

Efficiency of cycle
Efficiency of cycle (in percentage is defined as
`eta = ("Net W ork done in the cycle ")/("Heat absorbd in the cycle ")xx100`
or `eta = W_("total")/Q_("+Ve") xx100`
`=(Q_(+Ve) - Q _("-Ve"))/(Q_("+Ve")) xx100=(1 - Q_1/Q_2) xx100 " ...(i)"`
Where, `Q_1` negative heat in the cycle (heat released) and `Q_2` Positive heat in the cycle (heat absorbed) In the cycle
`Q_("AB") = Q_("CD") = 0` (Adiabatic process)
`Q_("DA") = nC_V Delta T=(1) (5/2R) (T_A -T_D)`
(` C_V = 5/2 R ` for a diatomic gas)
`(=5/2)xx8.31(300 - 7914)J" or "Q_("DA") = -10208.8J`
and `Q_(BC) = nC_P Delta T = (1)(7/2 R)(T_C -T_B), (C_P = 7/2 R "for a diatomic gas")`
`=(7/2)(8.331)(1818 - 909)J" or "Q_("BC") = 26438.3J`
Therefore, subsituting `Q_1 = 10208.8 J" and "Q_2 = 26438.3J` in
Eq. (i), we get `:." "eta ={1-(10208.8)/(26438.3)}xx100" or "eta = 61.4%`