Question

At 500 Kbar pressure density of diamond and graphite 3 gm/cc and 2gm/cc respectively at a certain temperature T .{delta H-deltaU} in KJ/ mole for the conversion of one mole of graphite to one mole of diamond at 500 Kbar pressure is 10^X .The value of X is ?

[given 1 bar =10⁵N/m²]

ans =2 how ?

Answer 1

Pressure = 500 K bar = 500,000 bar
Density = mass / volume.
Diamond and graphite are carbon. Hence, mass of 1 mole is 12 g.
Density of diamond = 3g/cc
Mass = 12g.
Hence volume of diamond = mass/density = 12/3 = 4 cc = 4×10⁻⁶ cubic meter (m³)
For graphite,
Density=2g/cc
Mass = 12 g
Volume = 12/6 = 2cc= 4×10⁻⁶ m³

Now;

We need
dH - dU.
We know,
H = U + PV
Here, constant pressure.
So,
dH = dU + PdV
dH-dU = PdV.
dV = change in volume = 4-2 = 2cc = 2×10⁻⁶ m³

Now,
dH-dU = PdV = 500,000 bar ×2×10⁻⁶ = 500,000 × 10⁵ N/m ×2×10⁻⁶ = 1000000 × 10 ⁻¹

= 10 ⁵
So answer is 5

But your answer says 2. I think this is an error in your book. If pressure was 500bar instead of 500Kbar, then, you will get 2 as the answer.

Either the question says 500 bar, or the answer given is incorrect in your textbook. Its a small error in calculation. Steps are however the same.