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What is the volume of 9.783x1023 atoms of He at 9.25 ATM and 512 K?
Updated: 9/20/2023
Wiki User
∙ 11y agoBest Answer
PV= nRT --> V=nRT/P
p = 9.25 atm
v= ?
n = 1.625
R= Raoult's Number (0.082057)
T= 512K
(1.6525*512*0.082057)/9.25= V
V= 7.506
Wiki User
∙ 11y agoThis answer is:
Anonymous
Lvl 1
∙ 3y agohow did you find n=1.635
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