If 1.20 moles of an ideal gas occupy a volume of 18.2 l at a pressure of 1.80 atm, what is the temperature of the gas, in degrees celsius?
-125°c
59.5°c
273°c
32°c

The equation that we will use to solve this problem is :
PV = nRT where:
P is the pressure of gas = 1.8 atm
V is the volume of gas = 18.2 liters
n is the number of moles of gas = 1.2 moles
R is the gas constant =  0.0821
T is the temperature required (calculated in kelvin)

Using these values to substitute in the equation, we find that:
(1.8)(18.2) = (1.2)(0.0821)(T)
T = 332.5 degree kelvin

The last step is to convert the degree kelvin into degree celcius:
T = 332.5 - 273 = 59.5 degree celcius