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

Answer:

$P_T=2.52atm$

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:

$P_2^{bulb\ 1}=\frac{2.00L*1.80atm}{5.00 L}=0.72atm\\\\ P_2^{bulb\ 2}=\frac{3.00L*3.00atm}{5.00 L}=1.8atm$

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!