What is the temperature of 0.257 mol of O2 occupying 6.78 L at 0.856 atm?

From the ideal gas law, PV = n RT

T = PV / n R

Given that n = 0.257 mole

V = 6.78 L

P = 0.856 atm

And R = gas constant = 0.0821 L. atm/K/mol

Therefore;

T = PV / n R

T = 0.856 atm * 6.78 L / 0.257 mole * 0.0821 L. atm/K/mol

= 275.1 K

=2.1 degree C