Answer:
[tex]\boxed{\text{60 kPa}}[/tex]
Explanation:
The volume and number of moles are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
[tex]\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}[/tex]
Data:
p₁ = 50 kPa; T₁ = 300 K
p₂ = ?; T₂ = 360 K
Calculation:
[tex]\begin{array}{rcl}\dfrac{50}{300} & = & \dfrac{p_{2}}{360}\\\\\dfrac{1}{6} & = & \dfrac{p_{2}}{360}\\\\\dfrac{360}{6} & = & p_{2}\\\\p_{2} & = & \mathbf{60}\\\end{array}[/tex]
[tex]\text{The new pressure will be $\boxed{\textbf{60 kPa}}$}[/tex]
