Solid Ammonium carbamate dissociates as:
`NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g).`
In a closed vessel, solid ammonium carbonate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of `NH_(3)` at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure. Also find the partial pressure of ammonia gas added.

Explanation

To solve the problem, we need to analyze the dissociation of solid ammonium carbamate and how the addition of ammonia affects the equilibrium. The dissociation reaction is given as:

\[ \text{NH}_2\text{COONH}_4(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g) \]

### Step 1: Define the initial conditions
Let’s assume the initial total pressure of the system is \( P \) atm when only solid ammonium carbamate is present. At this point, the partial pressures of ammonia and carbon dioxide are both zero since they are products of the dissociation.

### Step 2: Establish the equilibrium expression
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