Answer:
V₂ = 4.00 L
Explanation:
Given that:
Volume (v1) = 6.00 L
Temperature (T1) = 300 K
Pressure (P1) = 1.00 atm
VOlume (V2) = unknown???
Temperature (T2) = 600 K
Pressure (P2) = 3.00 atm
Using combined gas law equation:
[tex]\dfrac{P_1V_1}{T_1}= \dfrac{P_2V_2}{T_2}[/tex]
[tex]\dfrac{1 \times 6}{300} = \dfrac{3 \times V_2}{600}[/tex]
[tex]\dfrac{1}{50} = \dfrac{1 \times V_2}{200}[/tex]
200 = 50V₂
V₂ = 200/50
V₂ = 4.00 L
