What is the pressure inside a 35.0 L container holding 107.0 kg of argon gas at...continues
Homework Answers
n = 107/39.948 = 2.678
=> P = 2.678*0.0821*372/35
= 2.34 atm
.PV=nrT
=>P= n*.0821*372/37
n=107/40=2.675 kilo moles
P=2208 atmp note it's very high because mass of argon is in kgs
hence
PV = n RT
here n = 107/40 = 2.675 (molar mass of argon is 40)
R = 0.082 L atm/mol K
V = 35 L
T = 372
Hence P = 0.082 *372*2.675/35 = 2.33 atm
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