Answer:
The final temperature 271.77 °C
Explanation:
Given:
Initial pressure [tex]P_{1} =0.490[/tex] atm
Final pressure [tex]P_{2} = 0.655[/tex] atm
Initial temperature [tex]T_{1} =[/tex] 48° C = 321 K
Initial volume [tex]V_{1} = 6.38[/tex] L
Final volume [tex]V_{2} = 8.10[/tex] L
From ideal gas equation,
[tex]\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }[/tex]
[tex]T_{2} = \frac{P_{2} V_{2} }{P_{1} V_{1} } \times T_{1}[/tex]
[tex]T_{2} = \frac{0.655 \times 8.10 }{0.490 \times 6.38 } \times 321[/tex]
[tex]T_{2} = 544.77[/tex] K
[tex]T_{2} =[/tex] 271.77°C
Therefore, the final temperature 271.77 °C
