A334-ml cylinder for use in chemistry lectures contains 5.363 g of helium at 23 ∘c. you may want to reference (pages 404 - 407) section 10.4 while completing this problem. part a how many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior?

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)

We need to convert mL in L for the Ideal gas constant, because we have L as unit, so 334mL /1000 = 0,334 L

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

Look how the units are cancelled

(21,71 / 24,272) moles = n

0,894 moles = n

Moles . molar mass = gr.

0,894 moles . 8 g/m = 7,15 g

0.65 percent

Step-by-step explanation:

because 13/21 was doing it and therefor you divide 13 by 21 and get 0.65