The valve between the 2.00 L bulb, in which the gas pressure is 1.00 atm and the 3.00 L bulb, in which the gas pressure is 1.50 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant?

a. 0.900 atm
b. 1.25 atm
c.1.30 atm
d. 2.25 atm
e. 2.50 atm

Question

The valve between the 2.00 L bulb, in which the gas pressure is 1.00 atm and the 3.00 L bulb, in which the gas pressure is 1.50 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant?

a. 0.900 atm
b. 1.25 atm
c.1.30 atm
d. 2.25 atm
e. 2.50 atm

aaronq · Answer

Since the two bulbs are connected, now they are 1 system. But first we need to find their individual contributions (ignoring R and T because they're constants):

$P_1V_1=n_1$   and    $P_2V_2=n_2$             Add them up: $n_1+n_2=n_T$

Find the new pressure: $P_TV_T=n_T$


If you want you just substitute them in into 1 equation:  $P_T=\dfrac{P_1V_1+P_2V_2}{V_T}$