Answer:
1. Yes
2. No
3. Yes
Explanation:
Both containers have the same number of gas particles, so the moles, n, in the ideal gas law PV - nRT are both equal. So we can use the combined gas law will tell us which of the answers is possible.
P1V1/T1 = P2V2/T2, where P,V, and T are pressure, temperature (in K only) and volume. The subscripts are 1 for initial state and 2 for final state.
Lets say the top container is the initial state (P1,T1,V1), and the bottom container is the final state (P2,T2,V2).
P1V1/T1 = P2V2/T2 can be rearranged to tell us what might happen as we move from state 1 to state 2.
Pressure: P2 = P1(V1/V2)(T2/T1)
Volume: V2 = V1(T2/T1)(P1/P2)
Temperature: T2 = T1(P2/P1)(V2/V1)
We know from the diagram that V2> V1.
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Analyze the possible answers:
1. If the temperatures of both containers are
equal, container A has greater pressure than
container B. YES
Since P2 = P1(V1/V2)(T2/T1) and we know that a) V2 > V1, and b) T2 = T1, we can simplify and analyze what might happen. the equation reduces to P2 = P1(V1/V2)*1 since T2 = T1. We know that (V1/V2) < 1. Rearrange to (P2/P1) = (V1/V2). This means (P2/P1) < 1. Therefore P2 < P1. Container A has higher pressure than container 2. This is a possible outcome.
2. If the volume of container A decreased, its
pressure would decrease. NO.
Since this does not ask anything about container 2, gas states 1 and 2 are in the same container. So V1 = V2. There is no mention of temperature change, so T1 = T2. Lets use P2 = P1(V1/V2)(T2/T1)
P2 = P1(V1/V2)(T2/T1)
P2 = P1(V1/V2)(1) sine T2 = T1
Since V2 < V1, we can see that P1 will be greater than P2.
3. If the pressure in both containers is equal,
container A has a lower temperature than
container B. YES.
T2 = T1(P2/P1)(V2/V1)
P1 = P2 and V1 < V2, so we can write:
T2 = T1(1)(V2/V1)
Since V1<V2, the ratio (V2/V1) will be greater than 1. This means T2 will be greater than T1.