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an enclosed glass bulb of volume V contains a pinch of solid NH4Cl and 1/100 moles of Ne gas. At 300 K, the internal pressure inside the glass bulb is 114 mm Hg which increases to 908 mm Hg when the bulb is heated to 600 K.Assume ideal gas behaviour. Find the partial pressure of NH3 in the bulb at 600 K

```
3 years ago

Arun
25768 Points
```							At temperature 300 K, the internal pressure inside the bulb is solely due to the Ne gas, hence it follows the expression:P300K = (nNe RT) / V ----- (1)At temperature 600 K, the NH4Cl dissociates into NH3 and HCl, hence the total pressure inside the bulb will be due to the presence of Ne, NH3 and HCl gases, henceP600K = ((n1 + nNe) RT) / V ------(2)     [where n1= nNH3 + nHCl]Dividing expression (1) by (2) we get,114/908 = 0.01/(n1+ 0.01)n1 = 0.07Since NH4Cl dissociates into equal molar amounts of NH3 and HCl, the nNH3 and nHCl = 0.07 / 2 = 0.035 moles eachpartial pressure of NH3 = mole fraction of NH3x P600K PNH3 = 0.35 *908/0.35 +0.35 +0.01 = 397 mmHg
```
3 years ago
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