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is contained inside a cylinder under weightless piston.
The temepreture of ambient air is 27C and pressure is 1atm.

The mix inside the ballon is ignited and exploded, but cylinder was strong enough and remained undamaged. After some time thermal equilbruim was restored.


What is then new volume under the piston?

2007-07-23 11:31:23 · 4 answers · asked by Alexander 6 in Science & Mathematics Physics

4 answers

Let's say that initially there were 2/3 * N moles of H2 and 1/3 * N moles of O2, for a total of N moles of gas.

PV = NRT
N = PV / RT

After mixing, and returning to the state where the pressure and temperature are equal to ambient, there are now 2/3 * N moles of H2O, because
2*H2 + O2 = 2*H2O

We can calculate that
N = 12e-3 * 1.013e5 / 300 / 8.3143
= 0.487 moles of mixture before ignition
2/3 * N = 0.325 moles of H2 = 0.325 moles of water
water weighs 0.325 * 18 = 5.85 g
and assuming that the density of liquid water is 1 g/cm^3
this occupies ~ 5.85 cm^3

Note that 5.85 g was the mass of the H2 and O2 before mixing, and by mass conservation, that value cannot change after mixing.

2007-07-23 13:21:32 · answer #1 · answered by Dr D 7 · 1 0

Answer 1 is close but neglects a temperature-based correction of 273/(273+27), because the actual volume of a mole of gas is greater by the inverse of that ratio at 27C (22.4 L applying at STP, 0C), whereas liquid water's coefficient of expansion is negligible. So the answer is about 8.775 cc.

2007-07-23 12:20:57 · answer #2 · answered by kirchwey 7 · 0 0

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2017-01-21 14:43:53 · answer #3 · answered by jemmott 4 · 0 0

V = (12/22.4)*18gmw H20 = 9.643g ≈ 9.643 cc @ 27°C

2007-07-23 11:45:01 · answer #4 · answered by Steve 7 · 1 0

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