Answer:
[tex]\large \boxed{\text{104 kPa}}[/tex]
Explanation:
To solve this problem, we can use the Combined Gas Laws:
[tex]\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}[/tex]
Data:
p₁ = 748 kPa; V₁ = 273 m³; n₁ = n₁; T₁ = 525 K
p₂ = ?; V₂ = 1100. m³; n₂ = n₁; T₂ = 293 K
Calculations:
[tex]\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{748 kPa}\times \text{273 m}^{3}}{n _{1}\times \text{525 K}} & = &\dfrac{p_{2}\times \text{1100. m}^{3}}{n _{1}\times \text{293 K}}\\\\\text{390.0 kPa} & = &3.754{p_{2}}\\p_{2} & = & \dfrac{\text{390.0 kPa}}{3.754}\\\\ & = & \textbf{104 kPa} \\\end{array}\\\text{The new pressure is $\large \boxed{\textbf{104 kPa}}$}[/tex]
