OpenStudy (anonymous):

A student collects 530 mL of CO gas at a pressure of 1.2 atm and a temperature of 17ºC. What is the volume of the gas at 0ºC and 1 atm? (530 mL) x (1.2 / 1.0) x (273 / (17 + 273)) = 599 mL @Frostbite is this right?

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

how does it look @Frostbite?

4 years ago
OpenStudy (frostbite):

Please try calculate the number again.

4 years ago
OpenStudy (anonymous):

(530 mL) x (1.2 / 1.0) x (273 / (17 + 273)) is this part right?

4 years ago
OpenStudy (frostbite):

Yea it looks right, but can you try see what it gives again?

4 years ago
OpenStudy (anonymous):

ok

4 years ago
OpenStudy (anonymous):

598.717

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

wouldn't it be 590?

4 years ago
OpenStudy (frostbite):

wait a sec... switch the temperature fraction.

4 years ago
OpenStudy (anonymous):

i am so srry but I have to go grab dinner but I will be back in about 45 min. will u still be here?

4 years ago
OpenStudy (anonymous):

or will you be here tmrw?

4 years ago
OpenStudy (anonymous):

lol ur like my fave tutor :)

4 years ago
OpenStudy (anonymous):

yes my final answer is volume=676 is this right @Frostbite

4 years ago
OpenStudy (abb0t):

You're given all the information, I don't know why you didn't just use the ideal gas law formula: $$PV = nRT$$ to find moles. Then, Rearrange to solve for volume, therefore you have: $$V = \LARGE \frac{nRT}{P}$$

4 years ago
OpenStudy (anonymous):

:(

4 years ago
OpenStudy (anonymous):

wait so how do I use the PV=nRT formula?

4 years ago
OpenStudy (abb0t):

Did you change your temp to kelvin?

4 years ago
OpenStudy (abb0t):

I actually dont know i dont have a calculator with me so maybe ur right or wrong.

4 years ago
OpenStudy (anonymous):

yea

4 years ago
OpenStudy (anonymous):

ok :)

4 years ago
OpenStudy (anonymous):

thanks anyway!

4 years ago
OpenStudy (frostbite):

Sorry I said something stupid. I made a wrong arrangement between the gas laws.

4 years ago
OpenStudy (frostbite):

I'm so sorry, first now I see I did a fatal mistake: We don't actually need $$n$$, as it turns out it just go away: $\Large pV=nRT \to 1=\frac{ pV }{ nRT }$ $\Large \frac{ P _{1}V _{1} }{ nR T_{1} }=\frac{ P _{2}V _{2_{?}} }{ nRT _{2} }$ The gas constant $$R$$ and amount of substance $$n$$ remain constant so: $\Large \frac{ P _{1}V _{1} }{ T _{1} }=\frac{ P _{2}V _{2} }{ T _{2} }$ Ones again sorry, and thanks @abb0t you made me see it.

4 years ago
OpenStudy (abb0t):

I never said your method was incorrect. I wa just suggesting another :P

4 years ago
OpenStudy (frostbite):

Well it was incorrect :P

4 years ago