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 flasks, produced, final temperature, and final pressure in the flasks. (Assume `R = 0.08 L atm K^(-1) mol^(-1))`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the moles of NH₃ and HCl

**For NH₃:**
Using the ideal gas equation, \( PV = nRT \):
- \( P = 2.0 \, \text{atm} \)
- \( V = 1.0 \, \text{L} \)
...