The figure shows an insulated cylinder divided into three parts A,B and C. Pistons I and II are connected by a rigid rod and can move without friction inside the cylinder. Piston I is perfectly conducting while piston II is perfectly insulating. The initial state of the gas `(gamma =1.5)` present in each compartment A,B and C is as shown. Now , compartment A is slowly given heat through a heater H such that the final volume of C becomes `4V_(0/9)`. Assume the gas to be ideal and find.

Options

A. Final pressures in each compartment A,B and C
B. Final temperatures in each compartment A,B and C
C. Heat supplied by the heater
D. Work done by gas in Aand B
Heat flowing across piston I.

Answer

(a)finla pressure in `A=27/8P_0=`Final pressure in C, finla pressure in `B=21/4P_0`
(b)finla temperature in (AandB)`=21/4 t_0`,finla temperature in `C=3/2T_0`,
(c) `18 P_0V_0`,
(d)Work done by gas in `A= +P_0V_0`Work done by gas in B=0,
`(e)(17)/(2)P_0V_0`

Explanation

(a)For compartment -C
`PV^(r)=constant`
`P_(0)V_(0)^(3/2)=P_(1)((4V_(0))/(9))^(3/2)`
`P_(1)=P_(0)(9/4)^(3/2)=(27p_(0))/(8)`
`P_(1)=` Final pressure in A = finla pressure in c as movable pistion is in equilibrium Temperature in compartment A
`(P_(0)V_(0))/(T_(0))=(P_1((14V_0)/(9)))/(T_(0))rArr(P_0V_0)/(T_0)=((27)/(8)) (P_(0)xx((14)/(9))V_(0))/(T)`
Temperature in A = Temperture in B(as conducting) For C
`(P_0V_0)/(T_0) = (4V_(0))/(9)xx(27P_(0))/(8T_(c)`
`T_(c) = 4xx(27)/(9xx8)T_(0)=(3)/(2)T_(0)`
pressure in 'B' = P_(B)`
Then `
`(P_(B))/(T)=(P_(0))/(T_(0))rArrP_(B)=(P_(0)T)/(T_(0))=(21)/(4)(T_0P_0)/(T_(0))=(21)/(4)P_(0)`
Heat supplical by heat r `= Q_(b)`
`triangleQ_(b)=nC_(v)triangle(T)=(f)/(2)(nRT_(2)-nRT_(1))`
`=(f)/(2)(V_(0))(P_(B)-P_(0))`
`(f)/(2)V_(0)((17)/(4)P_(0))`
`=(17fP_(0)V_(0))/(8)` ltbRgt `r=1.5rArrf=4`
`Qtriangle_(b)=(17)/(2)P_(0)V_(0)`
ForA"
`P_(A)=P_(c)`
`V_(A)=W_(0)-V_(c)`
`dV_(A)=dV_(c)`
`W_(A)=intP_(A)dV_(A)-intP_(c)dV_(c)`
`=-W_(c)`
`=[(P_(i)V_(i)-P_(f)V_(F))/(R_(1))]`
`=[(P_(f_c)-P_(i_c)V_(i_c))/(r_(1))]`
`W_(A)=((4V_(0))/(9)xx(27P_(0))/(8)-P_(0)V_(0))/(1//2)`
`=(P_(0)v_(0))/(1//2)=P_(0)V_(0)`
`triangleU_(A)=(f)/(2)(nRtriangleT_(A))=(P_(1)xx(14V_(0))/(9)-P_(0)V_(0))f/2`
`=((27)/(8)P_(0)xx(14V_(0))/(9)-P_(0)V_(0))(f)/(2)`
`((21)/(4)P_(0)V_(0)-P_(0)v_0)(f)/(2)`
`=((17)/(4)P_(0)V_(0))xx2`
`=triangleQ=(17)/(4)P_(0)V_(0)+(17)/(2)P_(0)v_(0)`
`=18P_(0)V_(0)`
`W_(a)=P_(0)V_(0)`
`Heat flow = triangleQ_(b)=(17)/(2)P_(0)V_(0)`