Question

At 300.0 K and 0.987 atm pressure, what will be the volume of 2.30 mol of Ne?

Answer 1

V = 57.39 L

Step-by-step explanation:

Given that,

Temperature, T = 300 K

Pressure, P = 0.987 atm

No. of moles of Ne, n = 2.30 mol

We need to find the volume of Ne. We know that, the ideal gas law is as follows :

PV = nRT

Where

P is pressure and R is gas constant

`V=(nRT)/(P)\\\\V=(2.3* 0.0821* 300)/(0.987 )\\\\V=57.39\ L`

So, the volume of the Ne is 57.39 L.