Question

A flexible container at an initial volume of 3.10 l contains 7.51 mol of gas. more gas is then added to the container until it reaches a final volume of 11.9 l. assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Answer 1

Number of moles added: 21,319

Further explanation

There are several gas equations in various processes:

• 1. The ideal ideal gas equation

$\rm PV = nRT$

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 = mole

No = Avogadro number (6.02.10²³)

n = m / m

m = mass

M = relative molecular mass

• 2. Avogadro's hypothesis

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

$\rm V1: V2 = n1: n2$

• 3. Boyle's Law

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

$\rm p1.V1 = p2.V2$

• 4. Charles's Law

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

$\rm \dfrac {V1} {T1} = \dfrac {V2} {T2}$

• 5. Gay Lussac's Law

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

$\rm \dfrac {P1} {T1} = \dfrac {P2} {T2}$

• 6. Law of Boyle-Gay-Lussac

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

$\rm \dfrac {P1.V1} {T1} = \dfrac {P2.V2} {T2}$

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

The initial condition of A flexible container

initial volume: 3.10 l contains 7.51 mol of gas

final volume: 11.9 l

we use Avogadro's hypothesis

V1: V2 = n1: n2

$\rm \dfrac{3.10}{11.9}=\dfrac{7.51}{n_2}\\\\n_2=28,829$

Number of moles added:

 28,829-7.51 = 21,319 mol

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Answer 2

The  number  of  moles  of  a  gas  added to the container is  calculated   as  follows
intial  concentration= 7.51 moles
final  moles  - initial  moles

final  moles  icalculation
3.10L  = 7.51 moles
11.9L =?   moles
by  cross  multiplication

11.9  L x  7.51 moles/  3.10 L = 28.83 moles

the  number  of moles added is therefore=  28.83  moles -7.51  moles = 21.32 moles