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.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

a.   mass number from atomic number