Answer:
[tex]\boxed{\text{2.6 kPa}}[/tex]
Explanation:
To solve this problem, we can use the Combined Gas Laws:
[tex]\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}[/tex]
Data:
p₁ = 1.7 kPa; V₁ = 7.5 m³; T₁ = -10 °C
p₂ = ?; V₂ = 3.8 m³; T₂ = 200 K
Calculations:
(a) Convert temperature to kelvins
T₁ = (-10 + 273.15) K = 263.15 K
(b) Calculate the pressure
[tex]\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}[/tex]
