Question

Given one mole of diamond vs one mole of graphite,

Answer 1

Answer:

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

The temperature is $T =1218.63 K$

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

          $P = \frac{TS-E}{V}$

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

                              $3.42cm^3 = \frac{3.42}{1*10^6} = 3.42*10^{-6}m^3$ for V and $2.38 J/K$ for S

     $P = \frac{(300 * 2.38)- 2900}{3.42*10^{-6}}$

         $P= - 6.39*10^8Pa$

The negative sign signifies the direction of the pressure

Given that  $P = 1*0^5Pa$

making T the subject

            $T = \frac{PV+E}{S}$

Substituting into the equation

            $T = \frac{1*10^5 * 3.42 *10^{-6}+2900}{2.38}$

                $T =1218.63 K$

             

         

Answer 2

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

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