$$PV = nRT$$
where:
* P is the pressure of the gas in atmospheres (atm)
* V is the volume of the gas in liters (L)
* n is the number of moles of gas
* R is the ideal gas constant (0.08206 L atm / mol K)
* T is the temperature of the gas in Kelvin (K)
In this case, we have:
* n = 7.81x10-3 mole
* V = 0.355 L
* T = 398 K
Substituting these values into the ideal gas equation, we get:
$$P = \frac{nRT}{V}$$
$$P = \frac{(7.81x10-3 mole)(0.08206 L atm / mol K)(398 K)}{0.355 L}$$
$$P = 7.16 atm$$
Converting atm to kPa:
P = 7.16 atm * 101.325 kPa/ atm = 728 kPa
Therefore, the pressure exerted by the oxygen gas is 728 kPa.
