Question

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

Answer 1

Answer:

V = 57.39 L

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

$V=\dfrac{nRT}{P}\\\\V=\dfrac{2.3\times 0.0821\times 300}{0.987 }\\\\V=57.39\ L$

So, the volume of the Ne is 57.39 L.