Asked by Ana
0.75 mol of argon gas is admitted to an evacuated 40 cm^{3} container at 40 C. The gas then undergoes an isochoric heating to a temperature of 500 C.
Before I forgot to write the question...so the question is What is the final pressure of the gas?
* physics - drwls, Friday, December 11, 2009 at 11:07am
What is the question?
I had to look up "isochoric". It means "contant volume". During that heating process, p will increase such that it remains proportional to (absolute) T. The absolute temperature increases by a factor 773/313 = 2.47. Pressure increases by the same factor. They do not tell you the initial pressure, but you can get the pressure using
P = nRT/V
where R = 82.06 cm^3*atm/mole K
Po = 0.75*82.06*313/40 = 482 atm
I have to get the answer in kPa so I got 4.883865e^4 kPa...but it is wrong
Before I forgot to write the question...so the question is What is the final pressure of the gas?
* physics - drwls, Friday, December 11, 2009 at 11:07am
What is the question?
I had to look up "isochoric". It means "contant volume". During that heating process, p will increase such that it remains proportional to (absolute) T. The absolute temperature increases by a factor 773/313 = 2.47. Pressure increases by the same factor. They do not tell you the initial pressure, but you can get the pressure using
P = nRT/V
where R = 82.06 cm^3*atm/mole K
Po = 0.75*82.06*313/40 = 482 atm
I have to get the answer in kPa so I got 4.883865e^4 kPa...but it is wrong
Answers
Answered by
Ana
ok...now tired
P=(0.75)*(8.31)*(773-313)/(4*10^-5)=106373
106373/1000=106.37kPa..still wrong
P=(0.75)*(8.31)*(773-313)/(4*10^-5)=106373
106373/1000=106.37kPa..still wrong
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