Asked by Anna
Given that many laboratory gases are sold in steel cylinders with a volume of 43.8 L, What mass (in grams) of argon is inside a cylinder whose pressure is 17810kPa at 24∘C?
I've attempted this problem but I'm not getting the correct answer. Please explain if you can!
I've attempted this problem but I'm not getting the correct answer. Please explain if you can!
Answers
Answered by
Damon
p = 1.67*10^4 kilo Pascals
V = 43.8 L
T = 273 + 24 = 297 K
R - 8.314 Liters kilopascals/deg K
PV = n R T
n = (1.67*10^4)(43.8)/ [ 8.314*297 ]
n = 296 mols
296 * 40 grams/mol = 11840 grams
or 11.84 kg
V = 43.8 L
T = 273 + 24 = 297 K
R - 8.314 Liters kilopascals/deg K
PV = n R T
n = (1.67*10^4)(43.8)/ [ 8.314*297 ]
n = 296 mols
296 * 40 grams/mol = 11840 grams
or 11.84 kg
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