When the valve between the two containers is opened and the gases allowed to mix, how does the volume occupied by the n2 gas change? express your answer using two significant figures?

The volume occupied by the N₂ gas change : 5 L

There are several gas equations in various processes:

PV = NkT

N =number of gas particles

n = number of moles

R = gas constant (8,31.10 ^ 3 J / kmole K

k = Boltzmann constant (1,38.10 ^ -23)

n = = N / No

n = m / m

n = mole

No = Avogadro number (6.02.10 ^ 23)

m = mass

M = relative molecular mass

In the same temperature and pressure, in the same volume conditions, the gas contains the same number of molecules

So it applies: the ratio of gas volume will be equal to the ratio of gas moles

At a fixed temperature, the gas volume is inversely proportional to the pressure applied

When the gas pressure is kept constant, the gas volume is proportional to the temperature

When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature

Combined with Boyle's law and Gay Lussac's law

P1.V1 / T1 = P2.V2 / T2

P1 = initial gas pressure (N / m2 or Pa)

V1 = initial gas volume (m3)

P2 = gas end pressure

V2 = the final volume of gas

T1 = initial gas temperature (K)

T2 = gas end temperature

From the attached picture, it shows that if the stopvalve is opened then the two gas systems / containers will be interconnected (almost the same as the related vessel system). So that the total volume filled by the two gases (O₂ and N₂) becomes:

total volume = O₂ container volume + N₂ container volume

total volume = 3 L + 2 L

total volume = 5 L

a description of Charles’s law

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Charles's law

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State Boyle's, Charles's, and Gay-Lussac's laws

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Keywords: containers, valve, O₂, N₂, volume, gas

c. orbital shape i hope this is correct