The minimum number of tickets that could admit all of them is six (6).
This thing is impossible to explain in words, so I shall attempt it with a diagram:
Here are the six ladies:
( A ) ( B )
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( C ) ( D )
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( E ) ( F )
-- 'E' and 'F' are the daughters of 'C' and 'D' .
-- 'C' and 'D' are the daughters of 'A' and 'B' .
So look what we have now:
-- 'A' and 'B' are the mothers of 'C' and 'D' .
There's 2 of the mothers.
-- 'C' and 'D' are the mothers of 'E' and 'F' .
There's the OTHER 2 mothers.
-- 'A' and 'B' are the grandmothers of 'E' and 'F' .
There's the 2 grandmothers.
-- 'E' and 'F' are the daughters of 'C' and 'D' .
There's 2 of the daughters.
-- 'C' and 'D' are the daughters of 'A' and 'B' .
There's the OTHER 2 daughters.
You want to know what ? !
The group is even bigger than THAT.
There are also 2 GRAND-daughters in the family ... 'E' and 'F' .
So now you have a list of 12 people ! ... 4 mothers, 2 grandmothers,
4 daughters, and 2 grand-daughters ... and they all get in to the
Christmas Market with only six tickets. Legally !
Such a deal !
Don't forget : Christmas this year is also the first day of Chanukah !
All for the same price !
You have to divide the pressure exerted by the air into two partial pressures: of the dry air and of the water vapor. Combining these two values gives you the parameter.
The answer is 165.3 cm³.
P1 * V1 / T1 = P2 * V2 / T2
The initial sample:
P1 = 84.6 kPa
V1 = 215 cm³
T1 = 23.5°C = 23.5 + 273 K = 296.5 K
At STP:
P2 = 101.3 kPa
V2 = ?
T2 = 273 K
Therefore:
84.6 * 215 / 296.5 = 101.3 * V2 / 273
61.34 = 101.3 * V2 / 273
V2 = 61.34 * 273 / 101.3
V2 = 165.3 cm³
A. A child rubs a balloon
