Question

What is the temperature, in °C, of 0.938 moles of `Cl_2` gas with a pressure of 3.24 atm and volume of 9.31 L?

Answer 1

Using `PV = nRT`, about 392 degrees Kelvin or 119 degrees Celsius.

Explanation:

`PV = nRT`
P = pressure
V = volume in liters
n = moles
R = the universal gas constant, which is different depending on the unit of pressure
T = temperature in Kelvin

So for this question
P = 3.24 atm
V = 9.31 L
n = .938
R = .0821 `(L*atm)/(mol *K)`

`PV = nRT`
(3.24)(9.31) = (.938)(.0821)(T)
30.16 = 0.077T
T = 391.7 K

To convert Kelvin to Celsius, subtract 273.
391.7 - 273 = 118.7 C