A bicycle tyre is filled with air having pressure of 270kPa at 27 °...

Given here, ${P}_{1} = 270kPa, {T}_{1} = 27 + 273 = 300K, {T}_{2} = 36 + 273 = 309K$

Ideal gas equation is given by $PV = nRT$.

Considering isochoric process, i.e., constant volume so $P ∝ T$

Thus, $\dfrac{{P}_{1}}{{T}_{1}} = \dfrac{{P}_{2}}{{T}_{2}}$

$⇒ {P}_{2} = \dfrac{{P}_{1}}{{T}_{1}} \times {T}_{2} = \dfrac{270 \times \left(309\right)}{300}$

$= 278kPa$