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Consider 4.00 L of … - QuestionCove
OpenStudy (anonymous):

Consider 4.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.00 L and the temperature is increased to 36 ∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder. So, P2=_____mmHg.

1 year ago
OpenStudy (greatlife44):

we know the amount of moles is the same

1 year ago
OpenStudy (anonymous):

please help i think i am way off when i tried to work the problem

1 year ago
OpenStudy (greatlife44):

We know that pressure, volume and temperature are changing right? \[pV = nRT \] Essentially the changes in pressure temperature and volume are all happening where the number of moles is a constant. \[\frac{ pV }{ T } = nR \] now, we go like this, we're asked to find the new pressure, so this is the formula we get and must use for our problem. \[\frac{ p_{1}v_{1} }{ T_{1} } = \frac{ p_{2}v_{2} }{ T_{2} }\] now we're asked to solve for pressure 2. we can re-write our formula to solve for the second pressure. so this is what we get. \[\frac{ P_{1}V_{1}*T_{2} }{ T_{1}V_{2} } = P_{2}\] all we do is plug in what we're given.

1 year ago
OpenStudy (greatlife44):

@sistergrl4 don't worry i'll help you

1 year ago
OpenStudy (greatlife44):

take a look at what I wrote above

1 year ago
OpenStudy (anonymous):

ok so p1 is 4.00L and P2 is 2.00L

1 year ago
OpenStudy (anonymous):

and the V1 and V2 are the temperatures?

1 year ago
OpenStudy (greatlife44):

V = VOLUME P = pressure T = temperature

1 year ago
OpenStudy (anonymous):

ok so V1-4.00L, V2-2.00L and T1-20'C and T2-36'C

1 year ago
OpenStudy (greatlife44):

yes

1 year ago
OpenStudy (anonymous):

what about the gas

1 year ago
OpenStudy (greatlife44):

but remember @sistergrl4 we need to convert the temperature to Kelvins. to do that you need to add the temperature in C+273 = K

1 year ago
OpenStudy (greatlife44):

our temperatures are in celcius and it wont work out for this problem.

1 year ago
OpenStudy (anonymous):

oh ok

1 year ago
OpenStudy (anonymous):

do i add 20 and 36 to 273=K

1 year ago
OpenStudy (anonymous):

please work with me i feel kinda dumb because i havent done chemistry in a while

1 year ago
OpenStudy (greatlife44):

20+273 = Kelvins for temperature 1 36 + 273 = Kelvins for temperature 2

1 year ago
OpenStudy (anonymous):

1-293 K 2-309K

1 year ago
OpenStudy (greatlife44):

great

1 year ago
OpenStudy (greatlife44):

now we have T1 = 293K P1 = 0.48 atm V1 = 4.0 L T2 = 309k V2 = 2L P2 = ? So all you have to do now is enter all your information into this formula \[\frac{ P_{1}V_{1}T_{2} }{ T_{1}*V_{2} }\]

1 year ago
OpenStudy (anonymous):

OK FOR P1 V1 T2 DO I ADD OR MULTIPLY THE TOP NUMBERS

1 year ago
OpenStudy (greatlife44):

multiply :)

1 year ago
OpenStudy (anonymous):

OK I AM MULTIPLYING

1 year ago
OpenStudy (greatlife44):

\[\frac{ 0.48atm*4.0L*309k }{ 293K*2L } = P_{2}\]

1 year ago
OpenStudy (anonymous):

562.56/ 293*2L= 562.56/586=0.96p2

1 year ago
OpenStudy (greatlife44):

Almost :) I got something like 1.01 atm

1 year ago
OpenStudy (anonymous):

oh man

1 year ago
OpenStudy (anonymous):

what did i do wrong

1 year ago
OpenStudy (greatlife44):

remember we can convert this easily to mmHG by multiplying 1.01*760 so we'd get 767.6 mmHG

1 year ago
OpenStudy (anonymous):

thanks

1 year ago
OpenStudy (greatlife44):

anytime :)

1 year ago
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