Asked by Dre
A 334-mL cylinder for use in chemistry lectures contains 5.225 g of helium at 23 C. How many grams of helium must be releasedto reduce the pressure to 75 atm assuming ideal gas behavior?
Answers
Answered by
bobpursley
find grams at 75
PV=nRT solve for n, then convert to grams.
PV=nRT solve for n, then convert to grams.
Answered by
Damon
He is 4 grams/mol
so we have 5.225/4 = 1.31 mols in 0.334 Liter at 273 + 23 = 296 K
P V = n R T
R = 0.082 L atm deg K / mol
V = 0.334 L
n = 1.31 mol
T = 296
so
P = 1.31 * 0.082 * 296 / 0.334
= 95.2 atm
new P = 75
T R V the same
so P is proportional to n
75/95.2 = new n/ 1.31
new n = 1.03 mols
1.03 * 4 g/mol = 4.13 grams remain
5.225 - 4.13 = 1.1 grams released
so we have 5.225/4 = 1.31 mols in 0.334 Liter at 273 + 23 = 296 K
P V = n R T
R = 0.082 L atm deg K / mol
V = 0.334 L
n = 1.31 mol
T = 296
so
P = 1.31 * 0.082 * 296 / 0.334
= 95.2 atm
new P = 75
T R V the same
so P is proportional to n
75/95.2 = new n/ 1.31
new n = 1.03 mols
1.03 * 4 g/mol = 4.13 grams remain
5.225 - 4.13 = 1.1 grams released
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