Answer:
7,15 g of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior
Explanation:
If we assume an ideal gas behaviour we must use the general gas equation.
P.V = n R T
where P is pressure (65 atm)
where V is volume (0,334 L)
n the number of moles (in this case He2)
R the Ideal gas constant (0,082 L.atm/mol.K)
T is temperature in K (ºC + 273) (296K)
<u>We need to convert mL in L for the Ideal gas constant, because we have L as unit, so 334mL /1000 = 0,334 L</u>
65 atm . 0,334 L = n . 0,082 L.atm/mol.K . 296K
(65 atm . 0,334 L) / (0,082 mol.K/L.atm . 296K) = n
<em>Look how the units are cancelled</em>
(21,71 / 24,272) moles = n
0,894 moles = n
Moles . molar mass = gr.
0,894 moles . 8 g/m = 7,15 g