Answer:
[tex]\large \boxed{\text{5.8 L}}[/tex]
Explanation:
Data:
p₁ = 1 bar; V₁ = 5 L; T₁ = 0 °C
p₂ = 1 bar; V₂ = ?; T₂ = 45 °C
The pressure and the number of moles are constant, so, to calculate the volume, we can use Charles' Law.
[tex]\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}[/tex]
Calculations:
(a) Convert temperatures to kelvins
T₁ = (0 + 273.15) K = 273.15 K
T₂ = (45 + 273.15) K = 318.15 K
(b) Calculate the new volume
[tex]\begin{array}{rcl}\dfrac{V_{1}}{T_{1}} &= &\dfrac{V_{2}}{T_{2}}\\\\\dfrac{5}{273.15} &= &\dfrac{V_{2}}{318.15}\\\\0.018 &= &\dfrac{V_{2}}{318.15}\\\\{ V_{2}} &=& 0.018 \times 318.15\\&=& \textbf{5.8 L}\\\end{array}\\\text{The volume will be $\large \boxed{\textbf{5.8 L}}$}[/tex]
