Answer:
At 580.4 K of temperature will the gas reach a pressure of 825kPa
Step-by-step explanation:
If the gas is stored in a tank, at 273K (as initial temperature) let's apply Charles Gay Lussac law, where the volume doesn't change and the number of moles neither.
If the volume keeps on constant, pressure is been modified directly proportional to absolute T°.
P1 / T°1 = P2 / T°2
388kPa / 273K = 825kPa / T°2
(388kPa / 273K) . T°2 = 825kPa
T°2= 825kPa . 273K/388 kPa
T°2 = 580.4K