A 17.3 gram sample of krypton gas has a volume of 901 mil-liters at a pressure of 2.04 atm. The temperature of the Kr gas sample is ?

Answer 1

For this question, you will want to use the ideal gas equation, PV = nRT. The most important thing is to make sure all your units match and cancel out when you input them into the equation. If you aren't sure what units each term needs to be in, check the constant R, which has units L*atm/mol*K. To cancel units out, all the other terms must match the units of the constant, because you don't want to convert the constant's units.

P = pressure = 2.04 atm

V = volume = 901 mL x (1 L / 1000mL) = 0.901 L

n = amount of substance = 17.3 g Kr x (1 mol Kr / 83.798 g Kr) = 0.206 mol Kr

R = ideal gas constant = 0.0821 L*atm/mol*K

T = temperature = ?

Next, you will want to rearrange the equation to isolate the variable you are trying to find, in this case temperature.

Take PV = nRT and divide both sides by nR to get PV / nR = T

Then, solve for T

T = (2.04 atm x 0.901 L) / (0.0821 Latm/mol K x 0.206 mol)

T = 108.76 K

If you need the temperature in Celsius, you would subtract 273.15 to get -164.39C