Question

What pressure is exerted by a `8,60*mol` quantity of gas at a temperature of `320*K`, that is confined to a `16.0*L` volume?

Answer 1

`1430.678` Pa

Explanation:

Using ideal gas formula: `PV=nRT` .
Here, `V = 16L = 16 dm^3`
`n = 8.6 "moles"`
`R = 8.314`
`T = 47+273.15= 320.15" ""Kelvin"`

Hence `P = (nRT)/V = (8.6*8.314*320.15)/(16)`
`=1430.678 Pa`

Answer 2

Approx. `14*atm`.....

Explanation:

We use the Ideal Gas equation.....

`P=(nRT)/V=(8.60*cancel(mol)xx0.0821*(cancelL*atm)/(cancel(K)*cancel(mol))xx320*cancelK)/(16.0*cancelL)`

`=??*atm`....