Answer:
156 °C
Explanation:
This looks like a case where we can use Charles’ Law:
[tex]\dfrac{V_{1}}{T_{1}} =\dfrac{V_{2}}{T_{2}}[/tex]
Data:
V₁ = 3.45 L; T₁ = 282 K
V₂ = 5.25 L; T₂ = ?
Calculation:
[tex]\begin{array}{rcl}\dfrac{V_{1}}{T_{1}}& =&\dfrac{V_{2}}{T_{2}}\\\\ \dfrac{\text{3.45 L}}{\text{282 K}} &=&\dfrac{\text{5.25 L}}{T_{2}}\\\\\dfrac{\text{0.012 23}}{\text{1 K}} & = & \dfrac{5.25}{T_{2}}\\\\0.01223T_{2} & = & \text{5.25 K}\\T_{2} & = & \textbf{429 K}\\\end{array}[/tex]
(c) Convert the temperature to Celsius
T₂ = (429 – 273.15) °C = 156 °C
