6.00 liters of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergo the following processes: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the pressure, volume, and temperature of the gas in states 2 and \(3 .\) How many moles of the gas are there?

The given information is: - Initial state (state 1): \(P_1 = 3.00 ~\text{atm}\), \(V_1 = 6.00 ~\text{L}\), \(T_1 = 400 ~\text{K}\) - Process 1-2: Isothermal expansion (constant temperature) with \(V_2 = 4V_1\) - Process 2-3: Isobaric compression (constant pressure) - Process 3-1: Adiabatic compression to the original state

To find the number of moles, we will apply the ideal gas law first to the initial state: PV = nRT, and R is the ideal gas constant, 0.0821 \((\text{L} \cdot \text{atm}) / (\text{mol} \cdot \text{K})\) \(3.00 ~\text{atm} \cdot 6.00 ~\text{L} = n \cdot 0.0821 (\text{L} \cdot \text{atm}) / (\text{mol} \cdot \text{K}) \cdot 400. ~\text{K}\) Solving for n, we get: \(n = \frac{3.00 \cdot 6.00}{0.0821 \cdot 400} = \frac{18}{32.84} \approx 0.548 ~\text{moles}\)

Since the process from state 1 to state 2 is isothermal (constant T), the final temperature in state 2 is the same as the initial temperature, i.e., \(T_2 = T_1 = 400 ~K\). Now, we simply apply the ideal gas law in state 2: \(P_2V_2 = nRT_2\) Given \(V_2 = 4V_1 = 4 \times 6.00 ~\text{L} = 24.00 ~\text{L}\), we can solve for \(P_2\). \(P_2 = \frac{nRT_2}{V_2} = \frac{0.548 \times 0.0821 \times 400}{24.00} \approx 0.750 ~\text{atm}\) So, in state 2, \(P_2 = 0.750 ~\text{atm}\), \(V_2 = 24.00 ~\text{L}\), and \(T_2 = 400 ~K\).

Since the process from state 2 to state 3 is isobaric (constant P), \(P_3 = P_2 = 0.750 ~\text{atm}\). We can apply the ideal gas law in state 3: \(P_3V_3 = nRT_3\) We know that process 3-1 is adiabatic, so we can use the relation for adiabatic processes: \(P_1V_1^{\gamma} = P_3V_3^{\gamma}\) For a monatomic ideal gas, \(\gamma = \frac{5}{3}\). Now, we have \(P_1V_1^{\frac{5}{3}} = P_3V_3^{\frac{5}{3}}\). Given \(P_1 = 3.00 ~\text{atm}\), \(V_1 = 6.00 ~\text{L}\), and \(P_3 = 0.750 ~\text{atm}\), we can solve for \(V_3\): \(V_3 = \left(\frac{P_1}{P_3}\right)^{\frac{3}{5}}V_1 = \left(\frac{3.00}{0.750}\right)^{\frac{3}{5}} \times 6.00 ~\text{L} \approx 12.74 ~\text{L}\) Now, we can find \(T_3\) using the ideal gas law: \(T_3 = \frac{P_3V_3}{nR} = \frac{0.750 \times 12.74}{0.548 \times 0.0821} \approx 183.11 ~\text{K}\) So, in state 3, \(P_3 = 0.750 ~\text{atm}\), \(V_3 = 12.74 ~\text{L}\), and \(T_3 = 183.11 ~K\). The final results are: - Number of moles of gas: 0.548 moles - In state 2: \(P_2 = 0.750 ~\text{atm}\), \(V_2 = 24.00 ~\text{L}\), and \(T_2 = 400 ~K\) - In state 3: \(P_3 = 0.750 ~\text{atm}\), \(V_3 = 12.74 ~\text{L}\), and \(T_3 = 183.11 ~K\)