What is the temperature of 0.250 mol of neon in a 2.20 L container at 4.68 atm?
What is the temperature of 0.250 mol of neon in a 2.20 L container at 4.68 atm?
Homework Answers
Use ideal gas law to solve this.
PV = nRT
where P = pressure = 4.68 atm
V= volume = 2.20 L
n= number of mols of gas = 0.25 mol
gas constant R = 0.0821 l atm/mol/k
plug in these numbers to get temperature
T = PV/nR =4.68 atm*2.20 L/ 0.25 mol*0.0821 L atm/mol/k
= 501.63 K
Temperature = 501.63 K
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