Equilibrium pressure of calcium carbonate

[cheme2019]

Homework Statement

Calcium carbonate primarily occurs as two crystalline forms, calcite and aragonite. The value of∆!° for the transition
CaCO3(calcite) ⇌ CaCO3(aragonite)
is +1.04 kJ·mol-1 at 25°C. At that temperature the density for calcite is 2.710 g·cm-3, and that of aragonite is 2.930 g·cm-3. At what pressure will the two crystalline phases be at equilibrium at 25°C?

Homework Equations

ln(a_i)= ((molar V)/RT) * (P-1)
ΔG = -RT * ln(K_p
K_p = a_aragonite/a_calcite

The Attempt at a Solution

So I started by finding the molar volume of each by dividing the MW by the individual density and got
V_ara = 34.157 cm^3/mol and
V_cal = 36.930 cm^3/ mol
From here I used the activity equation to get that

a_i = e^(((molar V)/RT) * (P-1)))

And since

K_p = a_aragonite/a_calcite and
ΔG = -RT * ln(K_p

I can write that

ΔG = -RT * ln(a_aragonite/a_calcite)

This leads to

ΔG = -RT * (((molar V_ara/RT) * (P-1) - ((molar V_cal/RT)*(P-1))
Or
ΔG = - molar V_ara*(P-1)+(molar V_cal*(P-1))

Continuing
1040 J/mol =(-34.157 cm^3/mol ) * P + 34.157 cm^3/mol +(36.930 cm^3/mol) * P - 39.630 cm^3/mol

Solving for P I get 376 J/cm^3.
I am given that the answer is 3850.9 but no units and I wasn't able to convert my units to get that number...
I'm pretty sure my math is correct but I think I messed up on units somewhere. Any help would be greatly appreciated! Thanks!

 

[Bystander]

Step one: pick CGS or MKS; do not attempt to mix them when you're not comfortable with either; i.e., densities are expressed as kg/m³ in MKS, and as g/cm³ in CGS; pressures in force per unit area, not energy per unit volume ...

 

[cheme2019]

But can't you use energy per unit volume for pressure? Similar to how atm is J/L. Plus I'm only given ΔG in kJ/mol and I don't know how to convert that into CGS

 

[mjc123]

Atm is not J/L; 1 J/L = 1000 Pa; 1 atm = 101325 Pa