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