What is the Final Pressure?

Question

Two containers contain the same gas at different temps and pressures as indicated below

 

Large Container:  temp:200K        Pressure:2X10^5         Volume: 1L

 

Small Container:  temp: 600K      Pressure: 3X10^5          Volume: 2L

 

 

 

The two containers are then connected to each other using a thin tube, and the pressure and temp in both containers are allowed to equalize.  If the final temp is 300K, what is the final pressure?  Assume that the connecting tube has negligible volume and mass 

Answer 1

The pressure and temperatures will equalize.  We also know the volume of each container.

 

 

We can use the ideal gas law to compute the initial amounts of gas in each container: PV = nRT with R = 8.314 L·kPa/K·mol.

 

 

n_s = P₁V₁/RT₁ = (2 x 10⁵ Pa)(1L)/(8.314 L·kPa/K·mol)(200 K) = 0.120 moles

 

n_L = P₂V₂/RT₂ = (3 x 10⁵ Pa)(2L)/(8.314 L·kPa/K·mol)(600 K) = 0.120 moles

 

 

 

The gas in the two chambers can now be considered as a whole, so we have 2(0.120) = 0.240 moles in 3 liters of volume.  Using 300 K as the final temperature and the ideal gas law (again),

 

 

P_f = nRT_f/V_f = (0.240 mol)(8.314 L·kPa/K·mol)(300 K)/(3 L) = 200 kPa