Final Pressure in 5L & 10L Tanks: PV = nRT

  • Thread starter Thread starter gwiber
  • Start date Start date
  • Tags Tags
    Final Pressure
AI Thread Summary
The discussion focuses on calculating the final pressure in two gas tanks after opening a valve between them. The initial pressures are 911.925 kPa for the 5-liter tank and 4560 torr for the 10-liter tank. The final pressure is determined using the ideal gas law (PV = nRT) and weighted averages based on the volumes of the tanks. The weights for the average are derived from the respective volumes of the tanks. This method ensures an accurate equilibrium pressure is established at constant temperature.
gwiber
Messages
5
Reaction score
0
1. The valve between a 5 liter tank in which the gas pressure is 911.925 kPa and a 10 liter tank containing gas at 4560 torr is turned open and equilibrium is established at constant temperature. What is the final pressure in the two tanks in atmospheres?



2. PV = nRT



The Attempt at a Solution

 
Physics news on Phys.org
This is solved using weighted averages. The final pressure in the 2 tanks is the weighted average of the initial pressure. The weights come from the volumes: V1/(V1+V2) for tank 1, V2/(V1+V2) for tank 2.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top