If 3.25 g N2 gas is introduced into an evacuated 1.50 L flask at 325 K, what is the pressure inside the flask (in atm)?

2.06 atm

Explanation:

The computation of the pressure inside the flask is shown below;

As per the ideal gas law

we know that

P V = nRT

where

V = 1.50L

T = 325 K

R = 0.0821

Now moles of N_2 is

= 3.25 ÷ 28

= 0.116 mol

Now

P = NRT ÷ V

= 0.116 × 0.0821 × 325 ÷ 1.50

= 2.06 atm

im doing test rn im going with 2 but ill give answer if itss wrong,   : )