I have a problem stated as the following: $$\frac{p+B}{p_{atm}+B}=\left(\frac{\rho}{\rho_0}\right)^N\tag{1}$$ And I need to find the pressure at a depth of $10,000m$. We can do this by using the fact that: $$\frac{dp}{dh}=\rho g\tag{2}$$ Rearranging equation $(1)$ gives: $$\rho=\left(\frac{p+B}{p_{atm}+B}\right)^{1/N}\rho_0$$ And substituting this into $(2)$ will give: $$\frac{dp}{dh}=\left(\frac{p+B}{p_{atm}+B}\right)^{1/N}\rho_0g$$ Rearranging this and integrating gives: $$\int\frac{dp}{(p+B)^{1/N}}=\int\frac{\rho_0g\,dh}{(p_{atm}+B)^{1/N}}$$ $$\frac{N(p+B)^{\frac{N-1}{N}}}{N-1}=\frac{\rho_ogh}{(p_{atm}+B)^{1/N}}+C\tag{3}$$ Since we know that $p(h=0)=p_{atm}$ and so: $$C=\frac{N(p_{atm}+B)^{\frac{N-1}{N}}}{N-1}\tag{4}$$ Rearranging $(3)$ and substituting $(4)$ I get: $$p=\left[\frac{N-1}{N}\left(\frac{\rho_0gh}{(p_{atm}+B)^{1/N}}+\frac{N(p_{atm}+B)^{\frac{N-1}{N}}}{N-1}\right)\right]^{\frac{N}{N-1}}-B$$ We are given that:
- $N=7$
- $\rho_0=1000$
- $p_{atm}=1\times 10^5$
- $B=3000\times 10^5$
- $g=9.81$
My answer came out as $100\times 10^6Pa$ but the answer is $92.4\times 10^6Pa$. Can anyone see where I have gone wrong?