Help on chemistry gas law problem?

Question

A 4.0 liter sample of oxygen has a pressure of 95 kPa. At a temp of 198 degrees celcius the same amount of has occupies 6.0 liters and has a pressure of 101 kPa. What was the original temp of the gas?

PLEASE HELP AND ALSO SHOW STEPS OF HOW TO SOLVE THIS.... thank you.

Answer 1

For gas, always think about gas law!
You may worry that the pressure unit is kPa rather than atm. Not a issue!
Let the original gas temperature to be T K. We must have:
pv/t = nR = PV/T, where p,v,and t, and P, V, and T, represent two sets of Pressure, Volume, and Temperature, since n is the number of moles of oxygen and R is a gas constant. Both sides of the equation have exactly the same unit/s. That is:
6.0*101/(198+273) = 4.0*95/T
or: T = 4.0*95*(198+273)/(6.0*101) = 295 (K) = 22C.