Answer:
P₂= 141.15 kPa
Explanation:
Given that
Volume ,V= 3 L
V= 0.003 m³
Initial temperature ,T₁ = 273 K
Initial pressure ,P₁ = 105 kPa
Final temperature ,T₂ = 367 K
Given that volume of the cylinder is constant .
Lets take final pressure = P₂
We know that for constant volume process
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\P_2=P_1\times \dfrac{T_2}{T_1}\\P_2=105\times \dfrac{367}{273}\ kPa\\\\P_2=141.15\ kPa[/tex]
Therefore the final pressure = 141.15 kPa
P₂= 141.15 kPa
