A 0.65 mole quantity of O2 occupies 4.0 L at 20.◦C. What is the final pressure exerted by the gas?

Question

1. 0.92 atm
2. 3.9 atm
3. 6.2 atm
4. 1.1 atm
5. 0.27 atm

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Initially, I got .27 atm. I got this answer by plugging in the values into the equation [pv=nRT], coming out as [p*4=0.65*.082*20]. However, my assignment says this answer is wrong. May someone explain to me my error? Am I overlooking something? Thank you in advance~

Answer 1

You have to convert 20° C to Kelvin; you can't just plug it into the equation

PV=nRT

P=nRT/V=[(0.65moles)(0.082)(293.15K)]/4.0L

P=3.9 atm

Answer 2

All data dimensions must match the units of the Universal Gas Constant (R) being used... For your problem R = 0.08206 L-Atm/mole-K.

Answer 3

I see. Thank you!

Answer 4

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