At what temperature will 6.21 g of oxygen gas exert a pressure of 5.00 atm in a 10.0-l container?

PV = nRT
P = 5 atm
V = 10 L

molecular mass of oxygen gas = 16 X 2 =32 g/mole

so number of moles of oxygen gas =

6.21/32 = 0.194
so n = 0.194
R = 0.0821 L atm K^-1 mole^-1
T = ? K

5 X 10 = 0.194 X 0.0821 X T

50 = 0.0159 X T

T = 50/0.0159 = 3144.654 K

in degree c = 3144.654 - 273
= 2871.65 degree c

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