Convert pascal • cubic meter to ton of coal
Learn how to convert
1
pascal • cubic meter to
ton of coal
step by step.
Calculation Breakdown
Set up the equation
\(1.0\left(pascal \times cubic \text{ } meter\right)={\color{rgb(20,165,174)} x}\left(ton \text{ } of \text{ } coal\right)\)
Define the base values of the selected units in relation to the SI unit \(\left(joule\right)\)
\(\text{Left side: 1.0 } \left(pascal \times cubic \text{ } meter\right) = {\color{rgb(89,182,91)} 1.0\left(joule\right)} = {\color{rgb(89,182,91)} 1.0\left(J\right)}\)
\(\text{Right side: 1.0 } \left(ton \text{ } of \text{ } coal\right) = {\color{rgb(125,164,120)} 2.9288 \times 10^{10}\left(joule\right)} = {\color{rgb(125,164,120)} 2.9288 \times 10^{10}\left(J\right)}\)
Insert known values into the conversion equation to determine \({\color{rgb(20,165,174)} x}\)
\(1.0\left(pascal \times cubic \text{ } meter\right)={\color{rgb(20,165,174)} x}\left(ton \text{ } of \text{ } coal\right)\)
\(\text{Insert known values } =>\)
\(1.0 \times {\color{rgb(89,182,91)} 1.0} \times {\color{rgb(89,182,91)} \left(joule\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 2.9288 \times 10^{10}}} \times {\color{rgb(125,164,120)} \left(joule\right)}\)
\(\text{Or}\)
\(1.0 \cdot {\color{rgb(89,182,91)} 1.0} \cdot {\color{rgb(89,182,91)} \left(J\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 2.9288 \times 10^{10}} \cdot {\color{rgb(125,164,120)} \left(J\right)}\)
\(\text{Cancel SI units}\)
\(1.0 \times {\color{rgb(89,182,91)} 1.0} \cdot {\color{rgb(89,182,91)} \cancel{\left(J\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} 2.9288 \times 10^{10}} \times {\color{rgb(125,164,120)} \cancel{\left(J\right)}}\)
\(\text{Conversion Equation}\)
\(1.0 = {\color{rgb(20,165,174)} x} \times 2.9288 \times 10^{10}\)
\(\text{Simplify}\)
\(1.0 = {\color{rgb(20,165,174)} x} \times 2.9288 \times 10^{10}\)
Switch sides
\({\color{rgb(20,165,174)} x} \times 2.9288 \times 10^{10} = 1.0\)
Isolate \({\color{rgb(20,165,174)} x}\)
Multiply both sides by \(\left(\dfrac{1.0}{2.9288 \times 10^{10}}\right)\)
\({\color{rgb(20,165,174)} x} \times 2.9288 \times 10^{10} \times \dfrac{1.0}{2.9288 \times 10^{10}} = 1.0 \times \dfrac{1.0}{2.9288 \times 10^{10}}\)
\(\text{Cancel}\)
\({\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{2.9288}} \times {\color{rgb(99,194,222)} \cancel{10^{10}}} \times \dfrac{1.0}{{\color{rgb(255,204,153)} \cancel{2.9288}} \times {\color{rgb(99,194,222)} \cancel{10^{10}}}} = 1.0 \times \dfrac{1.0}{2.9288 \times 10^{10}}\)
\(\text{Simplify}\)
\({\color{rgb(20,165,174)} x} = \dfrac{1.0}{2.9288 \times 10^{10}}\)
Rewrite equation
\(\dfrac{1.0}{10^{10}}\text{ can be rewritten to }10^{-10}\)
\(\text{Rewrite}\)
\({\color{rgb(20,165,174)} x} = \dfrac{10^{-10}}{2.9288}\)
Solve \({\color{rgb(20,165,174)} x}\)
\({\color{rgb(20,165,174)} x}\approx3.4143676591 \times 10^{-11}\approx3.4144 \times 10^{-11}\)
\(\text{Conversion Equation}\)
\(1.0\left(pascal \times cubic \text{ } meter\right)\approx{\color{rgb(20,165,174)} 3.4144 \times 10^{-11}}\left(ton \text{ } of \text{ } coal\right)\)
