Question

You have a 39.49L sample of 8.483g of helium gas at 1.121atm. What is the expected temperature of this sample? Assume helium is an ideal gas under these conditions and report your answer in °C and in 2 decimal places.

Answer 1

Answer – We are given, mass of He = 8.483 g , volume = 39.49 L ,

P = 1.121 atm

First we need to calculate the moles of He

Moles of He = 8.483 g / 4.0026 g.mol^-1

                     = 2.12 moles

We know the Ideal gas law-

PV=nRT

So, T = PV/nR

          = 1.121 atm *39.49 L / 0.0821 L.atm.mol^-1.K^-1 *2.12 moles

          = 254.34 K

So, temp, T^oC = 273.34-273.15 K

                        = -18.81^oC