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
Answers
Answer:
The partial pressure of NH₃ in the bulb at 600 K is 397.25 mm Hg.
Explanation:
Given data:
An enclosed glass bulb contains a pinch of NH₄Cl and 1/100 = 0.01 moles of Ne
Pressure inside the glass bulb at 300 K = 114 mm Hg
Pressure inside the glass bulb at 600 K = 908 mm Hg
To find: the partial pressure of NH₃ in the bulb at 600 K
Let us consider
No. of moles of Ne be “n1” = 0.01 moles
No. of moles of NH₃ be “n2”
No. of moles of HCl be “n3”
Since at 300 K the internal pressure “P” of the bulb is fully due to Ne. So, according to the ideal gas law we can write the equation as
P at 300 K = (n1RT)/V
Or, 114 = (0.01RT)/V ….. (i)
where P, V and T are pressure, volume & absolute temperature and R is ideal gas constant.
Now, at 600 K the final internal pressure of the bulb is increased and is due to Ne, NH₃ & HCl, therefore
P at 600 K = (n1 + n2 + n3)RT / V
Or, 908 = (0.01 + n2 + n3)RT / V ….. (ii)
Dividing equation (i) by (ii), we get
114/908 = 0.01/ (0.01 + n2 + n3)
Or, (0.01 + n2 + n3) = 0.01 / 0.125 = 0.08
Or, n2 + n3 = 0.08 – 0.01 = 0.07
[∵ NH₄Cl separates into equal moles of NH₃ & HCl, so n2 = n3]
∴ 2 n3 = 0.07
∴ n3 = n2 = 0.07 / 2 = 0.035
The partial pressure of NH3 in the bulb at 600 K
= mole fraction of NH3 * P at 600 K
= [0.035 / (0.035+0.035+0.01)] * 908
= (0.035/0.08) * 908
= 397.25 mmHg
Answer:
Explanation:
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