An open vessel at 27°C is heated until 3/5 th of the air in it has been expelled . Assuming that the volume of the vessel remains constant find
(i) the air escaped out if vessel is heated to 900 K.
(ii) temperature at which half of the air escapes out.
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Answers
Answer
One should notice the fact that on heating a gas in vessel , there are the number of moles of gas which goes(vapours) out . the volume of vessel remains constant.
Let initial number of moles of gas at 300 K be 'n' .
On , heating 3/5 moles of air escape out a temperature T.
T = ( n – 3/5 n ) = 2n/5
Therefore, moles of air left at temperature .
(a) On heating vessel to 900K , let n1 moles be left ,
again n1T1 = n2T2
》n1 × 900 = 300 × n
》n = 1/3 n
》n = n/3
Therefore , moles escaped out at 900 K
n – n/3 = 2/n moles
(b) Let n/2 moles escape out at temperature T, then
As here at certain temperature, the moles of air quantity becomes half at vessel.
》 n1 × t1 = n2 × t2
》n/2 × T = n × 300
Here, we get temperature present in vessel = 600K
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