Answer:
819 K
Explanation:
Given data
- Initial pressure (P₁): 1 atm (standard pressure)
- Initial volume (V₁): 250 mL
- Initial temperature (T₁): 273.15 K (standard temperature)
- Final pressure (P₂): 0.500 atm
- Final volume (V₂): 1500 mL
- Final temperature (T₂): ?
We can find the final temperature using the combined gas law.
[tex]\frac{P_1 \times V_1 }{T_1} = \frac{P_2 \times V_2 }{T_2}\\T_2 = \frac{P_2 \times V_2 \times T_1 }{P_1 \times V_1} = \frac{0.500atm \times 1500mL \times 273.15K }{1atm \times 250mL}=819K[/tex]
