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.

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

is it a purebred? it looks more like a red nose, but it doesn't have the red could it be a staffordshire bull terrior? unsure if that's the right name for it sorry