shows a cylindrial tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the tempreture are `p_(1), T_(1)` on the left and `p_(2), T_(2)` on the right. the system is left for sufficient time so that the tempreture becomes equal on the two sides. (a) how much work has been done by the gas on the left part ? (b) find the final pressures on the two sides. (c ) find the final equilibrium tempreture. (d) how much heat has flown from the gas on the right to the gas on the left ?

Answer

A, B, C, D

Explanation

(a) Since the wall cannt be moved`
`thus, Delta U = 0`
`and Delta Q = 0`
`hence, Delta D = 0`
`(b)Let final pressure in LTS ` = p_1that in RHS = p_2
Since No. of moles remains constant.`
`So (p_1V)/(2RT_1) = (p_2V)/(2RT)`
`rArr p_1 = (p_1)/(T_1)`
` ` = (p_1T_1T_2(P_1+P_2)/(T_1lambda) `As T = (T_2T_1(P_1+P_2))/(lambda)`
`(p_1T_2(P_1+P_2))/(lambda)`
`Similarly `P_2 = (p_2T_2(P_1+P_2))/(lambda)`
`(c)Let` T_2gtT_1 and'T'`be the common temperature
Initialy `P-2v)/(2) = n_2RT_1`
`rArr n_1 = (P_1V)/(2RT_1)`
`Hense Delta Q = 0,Delta W = 0`
`Hense Delta U = 0`
` in case LHS,`Delta U_1 = 1.5n_2R(T-T_2)`
`in case LHS`
` = Delta U _2 = 1.5n_2R(T_2-T)`
`But Delta U_1-Delta U_2 = 0`
`rArr 1.5n_1R(T-T_1) =1.5n_2R(T_2-T) `
`rArr n_1T-n_1T_1 = n_2T_2-n_2T`
`rArrT(n_1+n_2) =n_1T_2+n_2T_2`
`rArr T = ( n_1T_1+n_2T_2)/(n_1+n_2)`
` = (P_1VxxT_1)/(2RT_1)+(P_2VxxT_2)/((2RT_2)/(P_1V)/(2RT_1)+(P_2V)/(2RT_1)`
` = ((P_1+P_2)T_1T_2)/(lambda)`
`as P_1T_2+P_2 = T_1lambda`
`(b)For RHS`
` Delta Q = Delta U = (asDelta W = 0)`
`1.5 n_2R(T_(2 )-T)`
` = (1.5 P_(2)V)/(2RT_(2)) R [T_(2) -((P_(1)+P_(2))T_(1)T_(2))/(P_(1)T_(2)+P_2T_(1))]`
` = (1.5P_(2)V)/(2RT_(2))xx(P_(1)T_(2)^2-P_(1)T_(1)T_(2))/(lambda)`
` = (1.5 P_(2) V)/(2RT_(2)) xx (T_(2) -P_(1) (T_(2)-T_(1)) / (lambda)`
` =(3P_(2)P_(1)V(T_(2)-T_(1)))/(4lambda)`