Question

After 273 m3 of ethylene oxide at 748 kPa and 525 K is cooled to 293 K, it is allowed to expand to 1100. m3. The new pressure is _____kPa

Answer 1

`\large \boxed{\text{104 kPa}}`

Step-by-step explanation:

To solve this problem, we can use the Combined Gas Laws:

`(p_(1)V_(1) )/(n_(1)T_(1)) = (p_(2)V_(2) )/(n_(2)T_(2))`

Data:

p₁ = 748 kPa; V₁ = 273 m³; n₁ = n₁; T₁ = 525 K

p₂ = ?; V₂ = 1100. m³; n₂ = n₁; T₂ = 293 K

Calculations:

`\begin{array}{rcl}(p_(1)V_(1))/(n_(1) T_(1)) & = & (p_(2)V_(2))/(n_(2) T_(2))\\\\\frac{\text{748 kPa}* \text{273 m}^(3)}{n _(1)* \text{525 K}} & = &\frac{p_(2)* \text{1100. m}^(3)}{n _(1)* \text{293 K}}\\\\\text{390.0 kPa} & = &3.754{p_(2)}\\p_(2) & = & \frac{\text{390.0 kPa}}{3.754}\\\\ & = & \textbf{104 kPa} \\\end{array}\\\text{The new pressure is $\large \boxed{\textbf{104 kPa}}$}`