Step-by-Step Solution
Step 1
We have the relation $1\;{\rm{Pa}} = 1\;{\rm{N/}}{{\rm{m}}^2}$, and the seal level pressure is $14.7\;{{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{i}}{{\rm{n}}^2}}}} \right. } {{\rm{i}}{{\rm{n}}^2}}}$.
We are required to find the value of $1\;{\rm{Pa}}$ in ${{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{f}}{{\rm{t}}^2}}}} \right. } {{\rm{f}}{{\rm{t}}^2}}}$, and the value of sea level pressure in ${{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{i}}{{\rm{n}}^2}}}} \right. } {{\rm{i}}{{\rm{n}}^2}}}$.
Step 2
The value of $1\;{\rm{lb}}$ is equal to the $4.4482\;{\rm{N}}$, and the value of $1\;{\rm{ft}}$ is equal to the $0.3048\;{\rm{m}}$.
Step 3
Now, convert the units.\[\begin{array}{c} 1\;{\rm{Pa}} = 1\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {{{\rm{m}}^{\rm{2}}}}}} \right. } {{{\rm{m}}^{\rm{2}}}}}\\ = 1\;\left( {\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}} \right) \times \left( {\frac{{1\;{\rm{lb}}}}{{4.4482\;{\rm{N}}}}} \right) \times {\left( {\frac{{0.3048\;{\rm{m}}}}{{1\;{\rm{ft}}}}} \right)^2}\\ = 2.08 \times {10^{ – 2}}\;{{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{f}}{{\rm{t}}^3}}}} \right. } {{\rm{f}}{{\rm{t}}^3}}} \end{array}\]
Step 4
Convert the given atmospheric pressure at sea level into Pascal.\[\begin{array}{c} 1\;{\rm{atm}} = 14.7\;{{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{i}}{{\rm{n}}^2}}}} \right. } {{\rm{i}}{{\rm{n}}^2}}}\\ = 14.7\;\left( {\frac{{{\rm{lb}}}}{{{\rm{i}}{{\rm{n}}^2}}}} \right) \times \left( {\frac{{4.4482\;{\rm{N}}}}{{1\;{\rm{lb}}}}} \right) \times {\left( {\frac{{1\;{\rm{in}}}}{{0.0254\;{\rm{m}}}}} \right)^2}\\ = 1.0135 \times {10^5}\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}} \times \left( {\frac{{1\;{\rm{Pa}}}}{{1\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}}}}} \right)\\ = 101.35 \times {10^3}\;{\rm{Pa}} \end{array}\]
Step 5
Convert the above pressure in kilopascal.\[\begin{array}{c} 1\;{\rm{atm}} = 101.35 \times {10^3}\;{\rm{Pa}}\\ = 101.35 \times {10^3}\;{\rm{Pa}} \times \left( {\frac{{1\;{\rm{kPa}}}}{{{{10}^3}\;{\rm{Pa}}}}} \right)\\ = 101.35\;{\rm{kPa}} \end{array}\]