Question

A 334-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?

Answer 1

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)

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

Answer 2

7.15 grams of helium must be released to reduce the pressure to 65 atm. assuming ideal gas behavior.

What is helium?

Helium is an element of the periodic table whose atomic number is 2.

It is the lightest gas of all.

By the ideal gas law

PV = nRt

where P = pressure which is 65 atm.

V = volume that is 334 ml or 0.334 L

n = number of moles

R is the gas constant

t = temperature, which is 296 K

Now, putting the values

$\rm 65 \times 0.334 = n \times 0.082 \times 296\\\\n = \dfrac{65 \times 0.334}{0.082 \times 296} = 0.894 \;mol$

Calculating the mass

0,894 mol × 8 g/m = 7.15 g.

Thus, the mass needed is 7.15 g.

Learn more about mass, here:

https://brainly.com/question/19694949