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Assume end correction approximately equa...

Assume end correction approximately equals to `(0.3) xx` (diameter of tube), estimate the approximate number of moles of ir present inside the tube (Assume tube is at `NTP`, and at `NTP, 22.4` litre contains `1` mole)

A

`(100pi)/(36 xx 22.4)`

B

`(10pi)/(18 xx 22.4)`

C

`(10pi)/(72 xx 22.4)`

D

`(10pi)/(60 xx 22.4)`

Text Solution

Verified by Experts

The correct Answer is:
A
End correction `= (0.3) d = 1 cm`
`d = (10)/(3) cm`
vol. of tube `= (pi(d^(2))/(4))l = (pi)/(4)((10)/(3))^(2) xx 100 cm^(2)` (take `I = 0.99 m =n 1 m`)
`= (10pi)/(36) lit`
moles `= (10 pi)/(36 xx 22.4)` moles (`22.4 "lt"`. contains `1` mole `(10pi)/(36)"lt"` contains `(10pi)/(36 xx 22.4)` mole)
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