Initially at 300.0 K and 10.1 atm pressure, 1 mol of Ar gas is allowed to expand isothermally against a constant pressure of 4.5 atm until equilibrium is reached. Assume van der Waals behavior with Cp,m = 28.58 + 1.76 x 10^-2 T (The units of T are K and the units of Cp,m are J K-1 mol-1.) Compute delta U, delta H, q, and w. Please explain each step in detail and the theory behind it
Question
Initially at 300.0 K and 10.1 atm pressure, 1 mol of Ar gas is allowed to expand isothermally against a constant pressure of 4.5 atm until equilibrium is reached. Assume van der Waals behavior with Cp,m = 28.58 + 1.76 x 10^-2 T (The units of T are K and the units of Cp,m are J K-1 mol-1.) Compute delta U, delta H, q, and w. Please explain each step in detail and the theory behind it
Solution
The problem involves an isothermal expansion of an ideal gas. In an isothermal process, the temperature of the system remains constant. This means that the internal energy (U) of the system also remains constant because internal energy is a function of temperature only. Therefore, the change in internal energy (ΔU) is zero.
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system (q) minus the work done by the system (w). Since ΔU = 0, we have q = w.
The work done by the system in an isothermal expansion against a constant external pressure can be calculated using the formula w = -P_ext(ΔV), where P_ext is the external pressure and ΔV is the change in volume. The negative sign indicates that work is done by the system.
The initial pressure (P1) is 10.1 atm, the final pressure (P2) is 4.5 atm, and the number of moles (n) is 1 mol. We can use the ideal gas law (PV = nRT) to calculate the initial and final volumes (V1 and V2). The gas constant (R) is 0.0821 L atm K-1 mol-1 and the temperature (T) is 300.0 K.
V1 = nRT / P1 = (1 mol)(0.0821 L atm K-1 mol-1)(300.0 K) / 10.1 atm = 2.44 L V2 = nRT / P2 = (1 mol)(0.0821 L atm K-1 mol-1)(300.0 K) / 4.5 atm = 5.47 L
The change in volume is ΔV = V2 - V1 = 5.47 L - 2.44 L = 3.03 L.
The work done by the system is w = -P_ext(ΔV) = -(4.5 atm)(3.03 L) = -13.64 L atm. To convert this to joules, we use the conversion factor 1 L atm = 101.3 J. So, w = -13.64 L atm * 101.3 J/L atm = -1381 J.
Since q = w, q = -1381 J.
The change in enthalpy (ΔH) for an ideal gas in an isothermal process is equal to the heat at constant pressure, which is q. So, ΔH = -1381 J.
In summary, ΔU = 0, ΔH = -1381 J, q = -1381 J, and w = -1381 J.
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