Physics 12.2 fhelp please

Question

0.61 mol of argon gas is admitted to an evacuated 90 mL container at 60 degrees C. The gas then undergoes an isochoric heating to a temperature of 500 degrees C. What is the final pressure of the gas? (in Pa)

Answer 1

Given:

n = 0.61 mol

V = 90 mL = 0.09L

T(i) = 60°C = 333K

T(f) = 500°C = 773K

Known:

R = 0.08214 (L*atm)/(mol*K)

Isochoric means isovolumetric, so the volume stays the same (V(i) = V(f), or ΔV = 0).

You can solve this the short way or the long way.

Short way:

P(f) = [n*R*T(f)] / V(f) = ...

Long way:

P(i) = [n*R*T(i)] / V(i) = ...

[P(i)*V(i)] / T(i) = [P(f)*V(f)] / T(f) --> P(f) = P(i) * {[V(i)*T(f)]/[V(f)*T(i)]} = P(i) * [T(f)/T(i)] = ...

Conversion factors for atmospheres (atm) and pascals (Pa):

1 atm = 1.013 × 10^5 Pa

or

1 Pa = 9.872 × 10^-6 atm