Question

The valve between the 2.00-L bulb, in which the gas pressure is 1.80 atm, and the 3.00-L bulb, in which the gas pressure is 3.00 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant

Answer 1

`P_T=2.52atm`

Step-by-step explanation:

Hello!

In this case, since we are analyzing the pressure-volume behavior of the gas, we need to use the Boyle's law as an inversely proportional relationship:

`P_1V_1=P_2V_2`

Thus, the final pressure in each bulb, once the gases have mixed, resulting in a volume of 5.00 L, is:

`P_2^(bulb\ 1)=(2.00L*1.80atm)/(5.00 L)=0.72atm\\\\ P_2^(bulb\ 2)=(3.00L*3.00atm)/(5.00 L)=1.8atm`

It means that the final total pressure, via the Dalton's law is:

`P_T=0.72atm+1.8 atm\\\\P_T=2.52atm`

Best regards!