Question

A 10.8g sample of gas has a volume of 5.25L at 27 degrees Celsius and 770mmHg. If 2.1g of the same gas is added to this constant 5.25L volume and the temperature raised to 61 degrees Celsius what is the new gas pressure? (give answer in mmHg)


My attempt

initial
p1=770mmHg
T1=27 degrees Cels.==> 300.15K
v1=5.25L
m1=10.8g

final
pf=?
Tf=61 degrees Cels.==>334.15K
vf=5.25L <--i think this was the problem, do they mean this stays constant?
mf= 12.9g <---I also think this is a problem I must have misunderstood it and added the initial mass to the amount of gas added to the 5.25L v

Answer 1

then I solved with
(pivi)/(niti)=(pfvf)/(nftf)
and isolated pf to get
pf=(pinftf)/(niti)

and subbed n=mass/molar mass

and the molar mass cancelled out

=(pimitf/molarmass)x(molarmass/m2ti)
and plugged in
pf =770mmHg(10.8g)(334.15K)/((12.9g)(300.15K))
=717.675mmHg

Answer 2

I'm not sure I understand the question...

Answer 3

your interpretation of the problem seems right to me

Answer 4

your solution is too

Answer 5

so your answer should be right

Answer 6

i know because pf < pi. that's right because you increased the temperature so that means pf is expected to be less than the initial pressure since temp and pressure are inversely proportional

Answer 7

okay I see thank you for replying I kept on submitting this answer but it was said to be incorrect