User:Theislikerice/Sandbox

How are Pascals and Joules related?

Pressure is force (units N) per area (units m^2)

${\mbox{Pa}}={{\mbox{N}} \over {\mbox{m}}^{2}}={\mbox{N}}{\mbox{m}}^{-2}$

Energy or work is force (units N) times length (units m)

${\mbox{J}}={{\mbox{N}}\times {\mbox{m}}}$

Taken together, we get:

${\mbox{Pa}}={{\mbox{J}} \over {\mbox{m}}^{3}}={\mbox{J}}{\mbox{m}}^{-3}$

Convert J into atm L

In SI units: ${\mbox{R}}=8.314{\mbox{J }}{\mbox{mol}}^{-1}{\mbox{K}}^{-1}$ $=8.314{\mbox{Pa }}{\mbox{m}}^{3}{\mbox{mol}}^{-1}{\mbox{K}}^{-1}$

Conversion factor for pressure:

$1={1{\mbox{atm}} \over 101,300{\mbox{Pa}}}$

Conversion factor for volume:

$1={1000{\mbox{L}} \over 1{\mbox{m}}^{3}}$

Calculation:

$1{\mbox{J}}=1{\mbox{Pa }}{\mbox{m}}^{3}={\mbox{Pa }}{\mbox{m}}^{3}{1{\mbox{atm}} \over 101300{\mbox{Pa}}}{1000{\mbox{L}} \over 1{\mbox{m}}^{3}}$

$={1000 \over 101300}{\mbox{atm }}{\mbox{L}}=0.00987{\mbox{atm }}{\mbox{L}}$

Applying this conversion factor to R:

${\mbox{R}}=8.314{\mbox{J }}{\mbox{mol}}^{-1}{\mbox{K}}^{-1}$ $=8.314{\mbox{J }}{\mbox{mol}}^{-1}{\mbox{K}}^{-1}{0.00987{\mbox{atm }}{\mbox{L}} \over 1{\mbox{J}}}$ $=0.08206{\mbox{ atm }}{\mbox{L }}{\mbox{mol}}^{-1}{\mbox{K}}^{-1}$

Reminder of fraction algebra

Dividing by a fraction:

$x\div {a \over b}=x\times {b \over a}$

A different version of the same rule:

${x \over {a \over b}}={x\times b \over a}$

Written as exponential:

${x \over {ab^{-1}}}={x\times b\times a^{-1}}$

Solving Arrhenius equation for k, T, E

Here is the Arrhenius equation relating rates at different temperatures:

$ln({k_{2} \over k_{1}})={E_{A} \over R}({1 \over T_{1}}-{1 \over T_{2}})$

To solve for activation energy

$ln({k_{2} \over k_{1}})={E_{A} \over R}({1 \over T_{1}}-{1 \over T_{2}})$

Multiply by R and divide by the expression in parenthesis

${R \over {1 \over T_{1}}-{1 \over T_{2}}}ln({k_{2} \over k_{1}})=E_{A}$

Then, switch the two sides of the equation

$E_{A}={R \over {1 \over T_{1}}-{1 \over T_{2}}}ln({k_{2} \over k_{1}})$

To solve for k2

$ln({k_{2} \over k_{1}})={E_{A} \over R}({1 \over T_{1}}-{1 \over T_{2}})$

Apply the inverse of ln() to the original equation, i.e. raise e to the power of LHS (left hand side) and RHS:

${k_{2} \over k_{1}}=e^{({E_{A} \over R}({1 \over T_{1}}-{1 \over T_{2}}))}$

...then multiply by k1:

$k_{2}=k_{1}e^{({E_{A} \over R}({1 \over T_{1}}-{1 \over T_{2}}))}$

To solve for T2

$ln({k_{2} \over k_{1}})={E_{A} \over R}({1 \over T_{1}}-{1 \over T_{2}})$

First, multiply by R and divide by the activation energy:

${R \over E_{A}}ln({k_{2} \over k_{1}})={1 \over T_{1}}-{1 \over T_{2}}$

Then, add ${1 \over T_{2}}$ and subtract the entire LHS:

${1 \over T_{2}}=({1 \over T_{1}}-{R \over E_{A}}ln({k_{2} \over k_{1}}))$

Now, take the reciprocal of the LHS and the RHS:

$T_{2}={1 \over {1 \over T_{1}}-{R \over E_{A}}ln({k_{2} \over k_{1}})}$

To solve for T1 or k1

Use the results above, but swap T1 and T2 at the same time as swapping k1 and k2.

Latex test

${\frac {-b-\operatorname {sqrt} \left(-4ac+b^{2}\right)}{2a}}$

1. $m={\frac {1}{2}}kg+{\frac {1}{4}}kg$
2. .
3. $m={\frac {1kg}{2}}+{\frac {1kg}{4}}$
4. .
5. $x={\frac {1}{2K}}+{\frac {1}{4K}}$