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Abundance and Isotopes

The densities...

Question

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^-3, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is

A. 9.92 × 10⁵

B. 9.92 × 10^{8

C. 9.92 × 10^7}

D. 9.92 × 10^{6

All options' unit is Pascal
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What should be the answer of this question according to you and why?
Please explain in detail.}

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Solution

Free energy change = Work done = PdV= P(V2-V1)
But density = mass/volume and volume= mass/density
assuming that the mass of carbon is 12g ( 1mole) we will get
1895J = P (12/2.25- 12/3.31)
so P = 1895J x (3.35x2.21)/ (3.31-2.25)x12 x10^6( pressure in atm is converted to pascal
=9.92x10^8 pascal (approximately)

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Similar questions

298 K

2.25

3.31 g c m^{- 3}

(Δ G^{0})

1898 J m o l^{- 1}

298 K

298 K

2.25

3.31

g {c m}^{- 3}

1895 J {m o l}^{- 1}

^{- 3}

(Δ G^{o})

^{- 1}

C (g r a p h i t e)

C (d i a m o n d)

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 k J m o l^{- 1}

2 \times 10^{- 6} m^{3} m o l^{- 1}

C (g r a p h i t e)

C (d i a m o n d)

C (g r a p h i t e)

C (d i a m o n d)

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}

1 b a r = 10^{5} P a

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 K J m o l^{- 1}

2 \times 10^{- 6} m^{3} m o l^{- 1}

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a

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