A flask of `1L` having `NH_(3)(g)` at `2.0atm` and `200K` is connected with the another flask of volume `800 mL` having `HCI(g)` at `8atm` and `200K` through a narrow tube of negligible volume. The two gases react to form `NH_(4)(CI(s)` with evolution of `43kJ mol^(-1)` heat. if heat capacity of `HCI(g)` at constant volume is `20 JK^(-1) mol^(-1)` and neglecting heat capacity of flask, `NH_(4)CI`, and volume of solid `NH_(4)CI` formed, calculated in the falscks, produced, final temperature, and final presure in the flasks. (Assume `R = 0.08 L atm K^(-1) mol^(-1))`

Explanation

`{:(,NH_(3)(g)+,HC1(g)rarr,NH_(4)C1(s),DeltaH=-43.0kJ,),(underset("mole")"Initial",(1xx2)/(0.08xx200),(8xx0.8)/(0.08xx200),,),("Final",=0.125,=0.4,0,),("mole",0,0.275,0.125,):}`
`:.` Heat produced `= 0.125 xx 43 = 5.375kJ`
The heat produced is used to increase the temperature of `HC1` left in flask since heat capacity of flask and `NH_(4)C1 = 0`
`:. Q = n xx C_(V) xx DeltaT`
`5.375 xx 10^(-3) = 0.275 xx 20 xx DeltaT rArr DeltaT = 977.27`
`:.` Final temperature =` 200 + 977.27 = 1177.27 K`
Also, final pressure
`= (nRT)/(V) = (0.275 xx 0.08 xx 1177.27)/(1.8) = 14.39atm`