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