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?

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Chemistry

Morales

29 January, 10:26

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?

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Marc Gould · Answer 1

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