Question

o what temperature in oC must a 0.500 mole sample of O2 gas be adjusted in order to achieve a pressure of 4.00 atm in a 4.25 L container? Report your answer to the nearest oC. (

Answer 1

141C

Step-by-step explanation:

We need to use the ideal gas law to solve this question. Ideal gas law: PV = nRT

P = pressure = 4.00 atm = 4.00 × 101.325 = 405.3 kPa (pressure needs to be converted to kilopascals)

V = volume = 4.25L

n = number of moles = 0.500

R = gas constant = 8.314 mol/K (found om chemistry data sheet)

T = temperature = ? unknown

Before any other steps are taken, we need to rearrange the the ideal gas law formula to make T the subject.

PV/nR = T

Now we can substitute the values.

(405.3 × 4.25) ÷ (0.500 × 8.314) = 414.3673322107 (answer in kelvin)

Convert 414.3673322107 Kelvin to Celsius.

0C = 273.15K (found on chemistry data sheet)

414.3673322107 - 273.15 = 141.2173322107C

141.2173322107 to nearest C = 141C