Question

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

Answer 1

@aaronq

Answer 2

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.

Answer 3

I have no idea what you mean @aaronq

Answer 4

Do you know what the question is asking?

Answer 5

nope

Answer 6

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

Answer 7

yea lol I just have a couple questions.

Answer 8

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.

Answer 9

50 is 323.15 and 25 is 298.15

Answer 10

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

Answer 11

In other words, what's the initial temperature?

Answer 12

so I divide 50/3.8 or 323.15/3.8

Answer 13

lets use some algebra to rearrange the whole equation first.

\(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\rightarrow \sf \dfrac{3.8~atm}{323.15~K}=\dfrac{P_2}{298.15~K}\)

\(\sf P_2=\dfrac{3.8~atm*298.15~\cancel K}{323.15~\cancel K}\)

Answer 14

now just carryout the operations and you should have the answer

Answer 15

got it thank youu!