Question

What will be the volume of `8g` of `H_2` gas at the pressure of `4atm` and temperature of `300K` ?

What will be the volume of `8g` of `H_2` gas at the pressure of `4atm` and temperature of `300K` .

`NOTE:`
take
`(H=1)`
`(R=0.082Lit-atm-mol^-1K^-1)`

Answer 1

`24.42=24420cm^3`

Explanation:

We are going to use the ideal gas law, which states that

`PV=nRT`

`P` is pressure in pascals `(Pa)`

`V` is volume in `m^3`

`n` is the number of moles of the substance

`R` is the gas constant

`T` is the temperature in Kelvin `(K)`

Since we are using atmospheric pressure as units and the gas constant as `0.082 Lit-atm-mol^-1K^-1`, then our volume will be in liters `(l)`.
So, plugging in `P=4atm`, `n="8g"/"2.016g/mol"~~3.97 "mol"`, `R=0.082`, and `T=300K`, we get

`V=(nRT)/P=(3.97*0.082*300)/4~~24.42l=24420cm^3` of hydrogen gas `(H_2)`.