One mole of monoatomic ideal gas undergoes a process `ABC` as shown in figure. The maximum temperature of the gas during the process `ABC` is in the form `(xPV)/(R)` . Find x .

Answer

`x=8`

Explanation

The maximum temperature of the gas will be during process `BC`, Process `BC` can be represented by straight line ,
y =mx+C
So `P = mV +C`
Putting point B & C gives `3P = 2mV + C…..(i)` `" "` `P=6mV + C …..(ii)`
So subtracting `2P = -4mV` So `m = -(P)/(2V)`
From (ii) `P = -(P)/(2V).6V + C implies C = 4P`
Hence we get equation as `y =(-(P)/(2V))x + 4P....(iii)`
where y is pressure and x is volume of gas . Putting y from above.Now we have
`xy = nRT implies (-(P)/(V)x^(2) + 4Px) = nRT....(iv)`
For maximum temperature `(dT)/(dx) = 0 implies (-(2P)/(2V)x + 4Px) = 0`
Hence `x= 4V` Putting in (iii)
We get `nRT_(max) = 2P(4V) = 8PV`
So `T_(max) = (8PV)/(nR) implies x=8`