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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
States of Matter

NTA Abhyas 2020 · Chemistry · States of Matter

A 2 L vessel is filled with air 50 ° C and a pressure of 3 atm. The...

A 2 L vessel is filled with air 50°50 ° C and a pressure of 3 atm. The temperature is now raised to 200 °C° C. A value is now opened so that the pressure inside drops to one atm. What will be the fraction of the total number of moles, inside escaped on opening the value? (Assume no change in the volume of the container).

Options

  1. A7.7
  2. B9.9
  3. C8.9
  4. D0.77

Correct answer

D. 0.77

Step-by-step solution

Given that, V=2L,T1=50+273=323K,P1=3atmV = 2L, {T}_{1} = 50 + 273 = 323K, {P}_{1} = 3atm
On heating V=2L,T2=200+273=473,P2=?V = 2L, {T}_{2} = 200 + 273 = 473, {P}_{2} = ?
Using P-T law, P1T1=P2T2\dfrac{{P}_{1}}{{T}_{1}} = \dfrac{{P}_{2}}{{T}_{2}}
P2=3×473323=4.39{P}_{2} = \dfrac{3 \times 473}{323} = 4.39 atm
and n=PVRT=3×20.0821×323=0.226n = \dfrac{PV}{RT} = \dfrac{3 \times 2}{0.0821 \times 323} = 0.226
Now value is opened till the pressure is maintained at 1 atm.
Thus, at constant V and T, P∝nP ∝ n
So 4.39∝0.2264.39 ∝ 0.226
So 1∝nleft1 ∝ {n}_{left}
So n=0.2264.39=0.052n = \dfrac{0.226}{4.39} = 0.052
So moles escaped out =0.226−0.052=0.174= 0.226 - 0.052 = 0.174
So fraction of moles escaped out =0.1740.226=0.77= \dfrac{0.174}{0.226} = 0.77
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