find grams at 75
PV=nRT solve for n, then convert to grams.
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?
2 answers
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
5 answers