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?
A)0.720 atm
B)2.28 atm
C)2.52 atm
D)1.80 atm
E)2.40 atm

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? 
A)0.720 atm 
B)2.28 atm 
C)2.52 atm 
D)1.80 atm 
E)2.40 atm

Quizplus · Accepted Answer

According to the combined gas law, when the valve between the two bulbs is opened, the pressure in the two bulbs will equalize. The combined gas law equation is:

(P1V1)/T1 = (P2V2)/T2

Since the temperature remains constant, we can simplify the equation to:

P1V1 = P2V2

We can plug in the values given in the problem to solve for P2:

(1.80 atm)(2.00 L) = P2(3.00 L)

P2 = (1.80 atm)(2.00 L) / 3.00 L = 1.20 atm

Therefore, the final pressure in both bulbs will be 1.20 atm.

The Valve Between the 2