One mole of an ideal mono-atomic gas is taken round cyclic process `ABC` as shown in figure below. Calculate work done

Answer

`(25)/8 (P_0 V_0)/R B`

Explanation

Maximum temperature of the gas will be somewhere between B and C. Line BC is a straight line. Therefore, P - V equation for the process can be written as
` p = -m V + c" "(y =m x + c)`
Here `m = (2 p_0)/V_0` and `C = 5 p_0`
Here `m = (2 p_0)/V_0 "and" c = 5 P_0`
` c = 5 p_0 " P= -((2p_0)/V_0)V + 5P_0`
Multiplying the equation by V,
`pV = -((2p0)/V_0)V^2 + 5p_0 V " (PV = RT for n = 1)"`
`"RT " = -((2p0)/V_0)V^2 + 5p_0V`
or `T = 1/R [5 p_0 V - (2p_0)/V_0 V^2]" ...(i)"`
For to be maximum, `(dT)/(dV) =0`
`rArr " "5p_0-(4p_0)/V_0 = V = 0 rArr V = (5V_0)/4 `
i.e., at `V = (5 V_0)/4`, (on line ), temperature of the gas is maximum. From Eq. (i) this maximum temperature will be:
`T_("m ax") = 1/R[5p_0 ((5V_0)/4) - (2p_0)/V_0 ((5V_0)/4)^2]`
`T_("m ax") = 25/8 (P_0 V_0)/R`