Answer:
[tex]P_2 = 7.33 \times 10^5 Pa[/tex]
Explanation:
As we know that the tyre is fully inflated
So here we can say that the volume will remain constant
So we have
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
so here we have
[tex]P_1 = 7 \times 10^5 Pa[/tex]
[tex]T_1 = 273 + 21 = 294 k[/tex]
[tex]T_2 = (273 + 35) = 308 k[/tex]
now we have
[tex]\frac{7 \times 10^5}{294} = \frac{P_2}{308}[/tex]
[tex]P_2 = 7.33 \times 10^5 Pa[/tex]