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Number of ideal gas molecules per $cm^3$

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The lowest pressure that is the best vacuum that can be created in a laboratory at 27 degree celsius is $10^{-11}$millimetre of Mercury at this pressure the number of ideal gas molecules per $cm^3$ will be ?

As said ideal gas so I applied Ideal gas equation
$ PV=nRT $
$[(10^{-11}×10^5)/(24.4×760)]×10^6$
this should give $n/V $
and multiplying it by avogadro number ahould give answer but it isn't ?

Symbols have there usual meanings

asked Mar 3, 2017 in Physics Problems by physicsapproval (2,290 points)
edited Mar 4, 2017 by physicsapproval
What answer do you get? What answer is given?.

The number of mm of Hg is missing from your question.
I have correced the question and also checked calculation
I am still not able to get the answer its given$ 3.22 × 10^5$ molecules/$cm^3$
I think it is the factor of $10^6$. This should be in the denominator, not the numerator.
okay  got it that was my mistake per $cm^3$ shoud be in denominator so $10^6 $ would be in denominator ; okay got my error :)

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