Question

Given one mole of diamond vs one mole of graphite,

G S V
1 mol Graphite 0J 5.74 J/K 5.30 cm^3
1 mol Diamond 2900J 2.38 J/K 3.42 cm^3
(Assume S and V are constant over the temperatures and pressures we’re considering, which is a reasonable approximation.)
(a) At room temperature 300 K, at what pressure does diamond become more stable?
(b) At standard pressure 10^5 Pa, at what temperature does diamond become more stable?

Answer 1

Correct Data:

G S V

1 mol Graphite 0 J 5.74 J/K 5.30cm³

1 mol Diamond 2900 J 2.38 J/K 3.42cm³

Answer:

a) P = -639 MPa

b) T =1218.63 K

Step-by-step explanation:

a) For stability of diamond, the gibb's free energy, G = 0 J

The Gibb's free energy is given by G = E + PV - TS

E + PV - TS = 0

where, E = enthalpy, P = pressure, S = entropy

E = 2900 J

T = 300 K

S = 2.38 J/K

V = 3.42 cm³ = 3.42 * 10⁻⁶ m³

2900 + P3.42 * 10⁻⁶ - 300(2.38) = 0

P3.42 * 10⁻⁶ = -2186

P = 2186/(3.42 * 10⁻⁶)

P = -639* 10⁶ Pa

P = -639 MPa

b) At P = 10⁵

E + PV - TS = 0

2900 + 10⁵(3.42 * 10⁻⁶) - T(2.38) = 0

T =1218.63 K

Answer 2

The pressure is
`P= - 6.39*10^8Pa`

The temperature is
`T =1218.63 K`

Step-by-step explanation:

Generally Gibbs free energy is mathematically represented as

`G = E + PV -TS`

Where E is the enthalpy

PV is the pressure volume energy (i.e PV energy)

S is the entropy

T is the temperature

For stability to occur the Gibbs free energy must be equal to zero

Considering Diamond

So at temperature of T = 300 K

`E + PV - TS = 0`

making P the subject

`P = (TS-E)/(V)`

Now substituting 300 K for T , 2900 J for E ,

`3.42cm^3 = (3.42)/(1*10^6) = 3.42*10^(-6)m^3` for V and
`2.38 J/K` for S

`P = ((300 * 2.38)- 2900)/(3.42*10^(-6))`

`P= - 6.39*10^8Pa`

The negative sign signifies the direction of the pressure

Given that
`P = 1*0^5Pa`

making T the subject

`T = (PV+E)/(S)`

Substituting into the equation

`T = (1*10^5 * 3.42 *10^(-6)+2900)/(2.38)`

`T =1218.63 K`