## anonymous one year ago In an industrial process, the temperature of a fixed amount of gas in a rigid container drops from 50 degrees Celsius to 25 degrees Celsius. What will be the final pressure in the container if the initial pressure is 3.8 atmospheres? 1.9 atm 3.5 atm 4.1 atm 7.6 atm

1. anonymous

@aaronq

2. aaronq

So for this question you would use: $$\sf \large \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$$ the temperature needs to be in kelvin though, so add 273 to the temps in celsius.

3. anonymous

I have no idea what you mean @aaronq

4. aaronq

Do you know what the question is asking?

5. anonymous

nope

6. aaronq

lol have you read it? do you know anything from this class..

7. anonymous

yea lol I just have a couple questions.

8. aaronq

So make an effort and show that you're trying. You need to follow the formula I posted earlier. It compares a closed system, at two instances, when the temp is 50 and 25 celsius.

9. anonymous

50 is 323.15 and 25 is 298.15

10. aaronq

great. we're also told that the initial pressure is 3.8 atm. At what temp ( 25 or 50) i this true?

11. aaronq

In other words, what's the initial temperature?

12. anonymous

so I divide 50/3.8 or 323.15/3.8

13. aaronq

lets use some algebra to rearrange the whole equation first. $$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\rightarrow \sf \dfrac{3.8~atm}{323.15~K}=\dfrac{P_2}{298.15~K}$$ $$\sf P_2=\dfrac{3.8~atm*298.15~\cancel K}{323.15~\cancel K}$$

14. aaronq

now just carryout the operations and you should have the answer

15. anonymous

got it thank youu!