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