Answer:
[tex]206.6\ \text{kPa}[/tex]
Explanation:
[tex]P_1[/tex] = Initial pressure = 200 kPa
[tex]P_2[/tex] = Final pressure
[tex]T_1[/tex] = Initial temperature = [tex]30^{\circ}\text{C}[/tex]
[tex]T_2[/tex] = Final temperature = [tex]40^{\circ}\text{C}[/tex]
We have the relation
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\\Rightarrow P_2=\dfrac{T_2}{T_1}P_1\\\Rightarrow P_2=\dfrac{40+273.15}{30+273.15}\times 200\\\Rightarrow P_2=206.6\ \text{kPa}[/tex]
The pressure that would be exerted after the temperature change is [tex]206.6\ \text{kPa}[/tex].
