Chemistry problem finding temperature

given
0.142 mol Argon
volume of Argon=834 ml Argon
Pressure of Argon=4.83

wanted: the temperature of the Ar gas sample on the celsius scale?

PV = nRT

(4.83 probably atmospheres) (0.834 Litres) = 0.142 mol (0.08206 L-atm/mol-K) (T)

T = 345.70 Kelvin
but rounded to 3 sig figs would be
T = 346 Kelvin

346 Kelvin - 273 = 73 Celsius

you can email me if the pressure was not atm's