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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

Yogiraj , 9 Years ago

Grade 12th pass

3 Answers

Sakshi

Last Activity: 9 Years ago

just refer to the definations of heat enthalipies of transformation and other enthalipies to get the answer.
u also need to use hess law for this question

Omgirish

Last Activity: 8 Years ago

Free energy = useful work done∆G = p∆v1895= P(v2 - v1)1895= P(12/2.25 - 12/3.31) × 10^-6(m) × . 10^-3(kg) .......v=m/d1895= P(5.33 - 3.62) × 10^-91895= P × 1.71 × 10^-9 P= 1895/1.71×10^-9 P= 1108 × 10^9 P= 1.1 × 10^6

Kushagra Madhukar

Last Activity: 5 Years ago

Dear student,

Please find the solution to your problem below.

Volume of Graphite, V_g = 12/2.25 cm3 = 12/2.25 x 10^-3 L

Volume of diamond, V_d = 12/3.31 cm3 = 12/3.31 x 10^-3 L

We know ΔG = – PΔV = – P x (V_g – V_d) x 101.3 [ Since, 1 L-atm = 101.3 J]

Hence, 1895 = – ( – 1.71 x 10^-3) x 101.3 x P

on simplifying we get, P = 10.93 x 10³atm = 11.08 x 10⁸ Pa

Hope it helps.

Thanks and regards,

Kushagra

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