Solve Physical Chemistry Problems: Pb(s, l) at 5000C

[cisidore]

I have these physical chemistry problems and I can't solve them, I would like to find some help on how to solve them.
Consider a gas whose (p, V, T) behavior can be described by the equation of state

P= nRT/V+ n2bRT/ V2, it is not an ideal gas!

Where b= 0.0515l mol-1 is a constant. The molar heat capacity at constant volume of the gas is independent of temperature and has a value 3R/2.
A number of moles n of such a gas are initially in a container of volume V1= 2l at a temperature T1= 300K and pressure p1= 9atm. The walls of the container are adiabatic. The clamps that keep one of the walls at a fixed position are released, and the wall moves against a constant pressure pex that is three times the value of the initial pressure p1 inside the container. The motion of the wall stops when equilibrium is achieved.

(a) Calculate the final temperature T2 of the gas.

(b) What is the molar heat capacity at constant pressure of the gas? Justify

(c) Calculate the change in the enthalpy H of the gas in the process.

(d) Calculate the work w in the process.

(e) Calculate the change in entropy of the gas S in the process.







Problem 2

The standard molar entropy of Pb (s) at 298.15K and 1 bar is 64.80J K-1 mol-1 . The molar heat capacity of Pb (s) at constant pressure of 1 bar is
Cmp (T) = 22.13+ 0.01172T + 1.00*105T2
Where the units of T and Cmp are, respectively, K and J K-1 mol-1. The melting point of lead at 1 bar is 327.4oC, and the molar enthalpy of fusion under these conditions is 4770J mol-1. The molar heat capacity at constant pressure of 1 bar of Pb(l) is
Cmp (T)= 32.51 – 0.00301T,
Where the units of T and Cmp are, respectively, K and J K-1mol-1.

Calculate the standard molar entropy of Pb(l) at 5000C.
Calculate delta G0 for the transformation
Pb(s, 298.15K, 1bar) ----Pb(l, 600.55K, 1 bar).

 

[Zefram]

I am happy to assist you with your physical chemistry problems. I will provide a step-by-step guide on how to solve each problem, but please note that I will not give you the final numerical values as it is important for you to practice solving the problems yourself.

Problem 1:

(a) To calculate the final temperature T2, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. However, since the gas in this problem is not an ideal gas, we need to use the given equation of state and solve for T2.

First, we need to find the initial number of moles n. Using the ideal gas law, we can calculate the number of moles as n = PV/RT1, where T1 is the initial temperature. Plugging in the values, we get n = (9atm * 2l)/ (0.0821 L atm mol^-1 K^-1 * 300K) = 0.0614 mol.

Next, we can use the given equation of state to solve for T2. Rearranging the equation, we get T2 = (P + n2bRT1/V1)/ (nR/V1 + n2bR/V1^2). Plugging in the values, we get T2 = (3*p1 + n2bRT1/V1)/ (nR/V1 + n2bR/V1^2).

(b) The molar heat capacity at constant pressure of the gas can be calculated using the formula Cp = Cv + R, where Cp is the molar heat capacity at constant pressure and Cv is the molar heat capacity at constant volume. The problem states that the molar heat capacity at constant volume is 3R/2, so we can substitute this value in the formula to get Cp = 3R/2 + R = 5R/2.

(c) The change in enthalpy H can be calculated using the formula H = nCp(T2 - T1). Plugging in the values, we get H = (0.0614 mol)(5R/2)(T2 - T1).

(d) To calculate the work w, we need to use the formula w = -pex(V2 - V1), where

 

[ravenprp]

First, it is important to understand the given equation of state and its parameters. The equation of state given is not for an ideal gas, but rather for a real gas. This means that the gas molecules have some volume and experience intermolecular forces, unlike an ideal gas which has negligible volume and no intermolecular forces. The constant b in the equation represents the volume occupied by one mole of the gas molecules.

(a) To calculate the final temperature T2 of the gas, we can use the ideal gas law: PV=nRT. Since the volume and number of moles are constant, we can rewrite the equation as P1/T1=P2/T2. Substituting the given values, we get T2= 900K.

(b) The molar heat capacity at constant pressure can be calculated using the formula Cp= (dH/dT)P, where dH is the change in enthalpy and dT is the change in temperature. Since the gas is expanding against a constant pressure, the change in enthalpy is equal to the work done, which is given by w= pex(V2-V1). Using the ideal gas law, we get w=pex(V2-V1)= pexnRT1/ pex= nRT1. Substituting the values, we get Cp= (nR/2)+(nR/2)= nR.

(c) The change in enthalpy can be calculated using the formula dH= nCp(T2-T1). Substituting the values, we get dH= nR(900K-300K)= 600nR.

(d) The work done in the process can be calculated using the formula w= pex(V2-V1). Substituting the values, we get w= 27nRT1.

(e) The change in entropy can be calculated using the formula dS= nCp ln(T2/T1)- nRln(V2/V1). Substituting the values, we get dS= nRln(900K/300K)- nRln(2)= 2nRln(3/2).

To solve the second problem, we need to use the equations for entropy and Gibbs free energy. The standard molar entropy of a substance can be calculated using the formula S= So+Cp ln(T/T0), where So is the standard molar entropy at a reference temperature