What will be the pressure of a sample of 48.0 grams of oxygen gas in a glass container of volume 5.2 L at 25 degrees Celsius?

What will be the pressure of a sample of 48.0 grams of oxygen gas in a glass container of volume 5.2 L at 25 degrees Celsius?
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Answer :

PV = nRTP = pressure = ?V = volume = 5.2 Ln = moles of gas = 48 g x 1 mole/32 g = 1.5 moles O2 gasR = gas constant = 0.0821 L-atm/K-moleT = temperature in K = 25 + 273 = 298 KSolving for P = nRT/V = (1.5)(0.0821)(298)/5.2P = 7.0 atm (to 2 significant figures)
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