S1. Use information from the appendices to predict whether diamond or graphite is the stable form of carbon at 1 atm pressure. If the density of graphite is 2.26 g/cm3 and that of diamond is 3.51 g/cm3 , to what value must you change the pressure to make diamond the stable form of carbon at room temperature

a) Graphite is slightly more stable than diamond at atmospheric conditions.

b) P = 14,998 atm

Explanation:

Given:

- The gibb's free energy of graphite formation ΔG_g = 0 KJ/mol

- The gibb's free energy of a diamond formation ΔG_d = 2.90 KJ/mol

- Density of graphite is p_g = 2.26 g/cm3

- Density of diamond is p_d = 3.51 g/cm3

- Molecular Mass of diamond M_d = 12 g / mol

- Molecular Mass of graphite M_g = 12 g / mol

Find:

Q)predict whether diamond or graphite is the stable form of carbon at 1 atm pressure.

Q)to what value must you change the pressure to make diamond the stable form of carbon at room temperature

Solution:

- The stability of Diamond at atmospheric conditions can be checked by looking at the Gibb's free energy of diamond at atmospheric conditions:

                                ΔG_d = 2.90 KJ/mol > 0

While that of graphite is:

                                ΔG_g = 0 KJ/mol

- Since graphite requires no extra energy at atmospheric conditions, it means its stable in its form. Whereas, diamond requires 2.90 KJ/mol amount of energy for formation. Hence, Graphite is slightly more stable allotrope of carbon.

- Finding the minimum pressure where the ΔG of formation for diamond is equal to zero. ΔG(P)_d = 0. We can use the ideal gas law to find this pressure as follows:

                       ΔG(P_2)_d - ΔG(P_1)_d = Δ V ( P_2 - P_1)

                       [(ΔG(P_2)_d - ΔG(P_1)_d) / Δ V ] = ( P_2 - P_1)

                       P_2 =  [(ΔG(P_2)_d - ΔG(P_1)_d) / Δ V ] + P_1

- For the expression above calculate Δ V , change in Volume for diamond and stable form graphite:

                         Δ V = V_diamond - V_graphite

                         Δ V = M_diamond / p_d - M_graphite / p_g

Plug in the values:

                         Δ V = (12 / 10^6*3.51) - (12 / 10^6*2.26)  

                         Δ V = -1.92*10^-6 m^3 / mol

- Use the Δ V change in volume and compute the pressure P_2:

                        P_2 =  [(0 - 2.90) / (-1.92*10^-6) ] + 1 atm

                        P_2 = (1.51*10^9)/(1.01*10^5) + 1 atm

                        P_2 = 14,998 atm

- The minimum amount of pressure required for diamond to be stable at room temperature is 15,000 times the atmospheric pressure.

answer; /// the above statement is (true);

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: