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, and volume of solid `NH_(4)CI` formed, calculate the final temperature, and final pressure in the flasks. (Assume `R = 0.08 L atm K^(-1) mol^(-1))`

Options

A. `5.375 J`
B. `4.375 k J`
C. `5.375 k J`
D. `6.375 J`

Answer

c

Explanation

`{:(,NH_(3(g)),+,HCl_((g)),to,NH_(4)Cl_((s)),DeltaH=-473.0 kJ),("Initial mole",(1xx2)/(0.08xx200),,(8xx0.8)/(0.08xx200),,),(,=0.125,,=0.4,,0),("Final mole",0,,0.275,,0.125):}`
`:.` Heat produce `=0.125xx43=5.375 k J`