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mr_godi [17]
4 years ago
15

A sealed, expandable container is initially at 2.000 L and 25°C. To what temperature must it be changed to have a volume of 6.00

L?
..I used V1/T1 = V2/T2 to get 75 degrees Celsius, but for some reason that's wrong. I'd love to know what I did wrong.
Chemistry
1 answer:
balandron [24]4 years ago
4 0

Answer:

The new temperature is 894 K or 621 °C

Explanation:

Step 1: Data given

Initial volume of the container = 2.000L

Initial temperature = 25.0 °C = 298 K

Volume increased to 6.00 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 2.00L

⇒with T1 = the initial temperature = 25 °C = 298 K

⇒with V2 = the increased volume 6.00 L

⇒with T2 = the new temperature

2.00 L / 298 K = 6.00 L / T2

T2 = 894 K = 621 °C

The new temperature is 894 K or 621 °C

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