At 20 l flash at 500 Kelvin ab gas 4 ATM the equilibrium passed a pressure of a 20.387 find the equilibrium pressure of a b

To solve the problem involving the equilibrium pressure of gas A at a certain state, we will break it down and apply the relevant concepts. Here are the steps:

1. Break Down the Problem

1. Identify the initial conditions:
   • Volume (V) = 20 L
   • Temperature (T) = 500 K
   • Initial pressure (P_initial) = 4 ATM
   • Pressure after equilibrium (P_equilibrium) = 20.387 ATM
2. Determine which gas law or principle will help us find the equilibrium pressure of gas B.

2. Relevant Concepts

We will use the ideal gas law, which states: $PV = nRT$ Where:

• $P$ = pressure (ATM)
• $V$ = volume (L)
• $n$ = number of moles of gas
• $R$ = ideal gas constant $(0.0821 \, \text{L} \cdot \text{ATM} / \text{K} \cdot \text{mol})$
• $T$ = temperature (K)

Since we are comparing two states of the same system, we can derive the equilibrium condition based on the change in pressure.

3. Analysis and Detail

Using the ideal gas law, we can express the number of moles before and after reaching equilibrium.

• For the initial state, we have: $n = \frac{P_{initial} \cdot V}{R \cdot T} = \frac{4 \cdot 20}{0.0821 \cdot 500}$ Calculating $n$: $n = \frac{80}{41.05} \approx 1.95 \, \text{mol}$

• Now, the equilibrium pressure will be determined based on the number of moles remaining at the new equilibrium state. The total pressure at equilibrium $P_{equilibrium}$ consists of both gases A and B.

Assuming no gas A is lost, but gas B has formed, thus we must find $P_B$ such that: $P_{equilibrium} = P_A + P_B$ Where $P_A$ contributes a certain value and we know: $P_{equilibrium} = 20.387 \, \text{ATM}$

If A remains at 4 ATM, then: $P_B = P_{equilibrium} - P_A = 20.387 - 4 \approx 16.387 \, \text{ATM}$

4. Verify and Summarize

To summarize:

1. Initial moles calculated $n \approx 1.95 \, \text{mol}$ at initial pressure.
2. At equilibrium, given $P_{equilibrium}$ and $P_A$, we find $P_B$.

Final Answer

The equilibrium pressure of gas B is approximately: $P_B \approx 16.387 \, \text{ATM}$