- #1

Samurai Weck

- 11

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## 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.