Q and R are two rigid containers of volume 3V and V respectively containing molecules of the same ideal gas initially at the same
temperature. The gas pressures in Q and R are p and 3p respectively. The containers are connected through a valve of negligible
volume that is initially closed.
The valve is opened in such a way that the temperature of the gases does not change. What is the change of pressure in Q?
A) +p
B) +p/2
C) -p/2
D) -P
The answer is B, but why?
Answers
1. The problem statement, all variables and given/known data Two rigid containers contain the same type of ideal gas and are connected by a thin tube with a valve. Container two is 3 times the volume of container 1. The initial pressures inside the containers are different, as are the temperatures. When the valve is opened, the temperatures inside the containers are maintained. Find an expression for the equilibrium pressure inside the containers.
2. Relevant equations PV = nRT 3. The attempt at a solution P1V1=n1RT1 and P2V2=n2RT2 But, V2=3V1 Total volume = V1 + V2, but V2 = 3V1 ΣV=(n1RT1)/P1+(n2RT2)/3P2 I decided to take the molar volume, as I can't calculate the number of moles, But i'm not sure if i can do this ΣV,m=(RT1)/P1+(RT2)/3P2 It's this step that really stumps me, because the temperature inside the two tanks are maintained at different values, so for this last step I took the average temperature.
P3ΣV,m=RT3 where P3 is the equilibrium temp and T3 is the average temperature of the two tanks. P3((RT1)/P1+(RT2)/3P2)=RT3 After some rearranging i come to the answer P3=T3/((T1/P1)+(T2/3P2))
so, here the right answer is (b) +p/2
Hope, it may help you.
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◆ Answer-
∆P(for Q) = p
◆ Explaination-
# Given-
P1 = p
P2 = 3p
V1 = 3v
V2 = v
# Solution-
As valve is opened to move freely, pressure on each side will be equal conserving total pressure.
P = (P1+P2)/2
P = (p+3p)/2
P = 2p
Change in pressure in container Q is
∆P = P-P1
∆P = 2p-p
∆P = p
Change in pressure for container Q is p.