A sample of ideal gas is taken through the cyclic process shown in the figure. The temerauture of the gas in state A is `T_(A)` =200 K. In states B and C the temperature of the gas is the same.

What is the greastes temperature of the gas during the cyclic process ?

Answer

`2400`

Explanation

At B and C `(T_(B)=T_(C))`
`(P_(B)V_(B))/T_(B)=(P_(C)V_(C))/T_(C)`
`(P_(B)V_(A))/T_(B)=(P_(C)3V_(A))/T_(C)`
`P_(B)=3P_(C)`
for lineAC P alphaV(straight line through origin)
so `P_(C)/P_(A)=V_(C)/V_(A)=(3V_(A))/V_(A)`
`Rightarrow P_(C)=3P_(A)`
thus `P_(C)=3P_(A) , P_(B)=3P_(C)=9P_(A)` .......(1)
`V_(C)=3V_(A),V_(B)=V_(A)`
`T_(A)=(P_(A)V_(A))/(nR)` .....(2)
from A to B sochoric P uparrow T uparrow so `T_(B) gt T_(A)`
for C to A, both (P,V)downarrow so T downarrow
Thus from B to C (we should have maximum temperature) `P=aV+b`
`RightarrowP=-((6P_(A))/(2V_(A)))+12 P_(A) Rightarrow P=-(3P_(A)V)/(V_(A))+12P_(A)`
`PV=nRT`
`(-(3P_(A)V)/V_(A)+12P_(A))V=nRT`
for `T_(max) (dT)/(dV)=0`
`(-6P_(A))/V_(A) V+12P_(A)=0`
`V=2V_(A) Rightarrow P=6P_(A)`
`T=(6P_(A)(2V_(A)))/(nR)=(12P_(A)V_(A))/(nR)`
`T_(max)=12T_(A)=2400K`