The densities of graphite and diamond at 298 ~K are 2.25 and 3.31 gcm^ -3 , respectively. If the standard free energy difference (Δ G^ ) is equal to 1895 ~J ~mol^ -1 , the pressure at which graphite will be transformed

Correct answer: D. none Δ G=-P Δ V aligned & 1895=P[((12)/(2.25)-(12)/(3.31)) ×10^ -6 ~m^3] & P=(1895 × 10^6)/(1.71)=1108.187 ×10^6 Bar & =1108.1871 ×10^ 11 ~Pa aligned

NEET 2003

Exam: NEET
Year: 2003
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2003 · Chemistry · Thermodynamics (C)

The densities of graphite and diamond at 298 ~K are 2.25 and 3.31...

The densities of graphite and diamond at

298 \mathrm{~K}

are 2.25 and

3.31 \mathrm{gcm}^{-3}

, respectively. If the standard free energy difference

\left(\Delta \mathrm{G}^{\circ}\right)

is equal to

1895 \mathrm{~J} \mathrm{~mol}^{-1}

, the pressure at which graphite will be transformed into diamond at

298 \mathrm{~K}

is:

Options

A $9.92 \times 10^8 \mathrm{~Pa}$
B $9.92 \times 10^7 \mathrm{~Pa}$
C $9.92 \times 10^6 \mathrm{~Pa}$
Dnone

Correct answer

D. none

Step-by-step solution

\Delta \mathrm{G}=-\mathrm{P} \Delta \mathrm{V}

\begin{aligned}& 1895=P\left[\left(\frac{12}{2.25}-\frac{12}{3.31}\right) \times10^{-6} \mathrm{~m}^3\right] \\ & P=\frac{1895 \times 10^6}{1.71}=1108.187 \times10^6 \mathrm{Bar}\\ & =1108.1871 \times10^{11} \mathrm{~Pa} \end{aligned}

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