Question

A tire has a pressure of 325 kPa at 10°C.

Answer 1

5 times that of initial pressure i.e 1625 kpa

Answer 2

Original temperature  =  10°C  =  283 K .
New temperature  =  50°C  =  323 K.

Factor of temperature increase  =  323/283 .

If volume is held constant, then pressure increases by the same factor.

         (325 kPa) x (323/283)  =  370.9... kPa  (rounded)