Using Ideal Gas Law to find P, V, or T

AI Thread Summary
The discussion revolves around using the Ideal Gas Law to calculate the final pressure of argon gas in a container after heating. The initial calculations for pressure were incorrectly assumed to yield results in atmospheres instead of pascals, leading to confusion in unit conversion. The correct approach involves recognizing that the pressure calculated using the ideal gas constant results in pascals, necessitating a conversion to kilopascals for the final answer. The user realized their mistake regarding the units and acknowledged the importance of understanding the relationship between the gas constant and pressure units. This clarification is crucial for solving similar problems correctly.
Samurai Weck
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Ok, so I'm doing this homework online and I THINK I'm doing it correctly, but I'm getting incorrect answers.

Homework Statement



0.290 mol of argon gas is admitted to an evacuated 40.0 cm^3 container at 60.0 degrees C. The gas then undergoes an isochoric heating to a temperature of 300 degrees C.

What is the final pressure of the gas?

Homework Equations



p=(nRT)/V and p(f)/T(f)= p(i)/T(i)

The Attempt at a Solution



Okay, so I need to find p(i).
n=.290 mol
V=40 cm^3 = 40cm^3(1m/100 cm)^3 = 4x10^-5 m^3
T= 60.0 degrees C = 60 C + 273 = 333 K
R= the ideal gas constant which is 8.31 J/mol K

Therefore,
p(i)= (.290 mol * 8.31 J/mol K * 333 K)/ (4x10^-5 m^3)
p(i) = 2.01x10^7 atm

Now that I know p(i), i can then solve for the temperature increase to find p(f)

p(f) = p(i)T(f)/T(i)
p(f) = (2.01x10^7 atm * 573 K)/ 333 K = 34586486 atm = 3.46x10^7 atm

They want the answer in kPa, and converting atm to kPa online yielded 3.50x10^10.

I'm completely clueless as to what I'm doing wrong. There are two other problems asking for either p or V which seem to have the same basis. If I can figure out what I'm doing wrong here, I should be able to understand why I'm doing the rest wrong. Thank you.
 
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Ah, disregard my question. I was mistaken when I thought that pressure in p=nRT comes out in atm. It actually is in Pa and you have to change 3.46x10^7 to 3.46x10^4.
 
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