Question

In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles of the gas are present in the tank? What is the molecular weight of the gas? Assuming that the gas to be a pure element can you identify it?

Answer 1

The molecular weight will be "28.12 g/mol".

Step-by-step explanation:

The given values are:

Pressure,

P = 10 atm

=
`10* 101325 \ Pa`

=
`1013250 \ Pa`

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

=
`1 \ m^3`

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒
`PV=nRT`

o,

⇒
`n=(PV)/(RT)`

By substituting the values, we get

`=(1013250* 1)/(8.3145* 298)`

`=408.94 \ moles`

As we know,

⇒
`Moles(n)=(Mass(m))/(Molecular \ weight(MW))`

or,

⇒
`MW=(m)/(n)`

`=(11.5)/(408.94)`

`=0.02812 \ Kg/mol`

`=28.12 \ g/mol`