Calculate the DECREASE in temperature when 6.00 L at 20.0 °C is compressed to 4.00 L ?

Question

I need help fast please help and it is about chemistry 11th grade So i like I need it now or soon please!!!

Answer 1

For this question you need one of the gas laws. You need Charles's law, which relates temperature and volume in the following way: V₁xT₁=V₂xT₂ where volume is in liters and temperature in Kelvin.

Here are the steps:

1. Convert the temperature to Kelvin using K = ^oC + 273
2. K = 20.0^oC + 273
3. K = 293 K
4. Plug in the initial volume, initial temperature, and final volume in the equation V₁xT₁=V₂xT₂
5. 6.00 X 293 = 4.00 X T₂
6. Solve for T₂:
7. 6.00 X 293 = 4.00 X T₂
8. 1758 = 4.00T₂
9. 1758÷4.00 = 4.00T₂÷4.00
10. 439.5 = T₂
11. The previous answer tells you what it will decrease to, so the total decrease will be the initial temperature - the final temperature:
12. 1758 - 439.5 =1318.5
13. The temperature decreases by 1318.5 K

Answer 2

You are given the following:

Volume 1 = V1 = 6.00 L

Temperature 1 = T1 = 20.0ºC +273 =293K

Volume 2 = V2 = 4.00 L

And you are asked for the new temperature (Temperature 2) = T2

Since no pressure is given, we will assume it to be constant.

At constant pressure you are dealing with Charles' Law: V1/T1 = V2/T2

Solving for T2 we have...

V2T1/V1 = T2

T2 = (4.00L)(293K)/6.00L)

T2 = 195K