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:

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

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


 * 1) $$ m = \frac{1}{2} kg + \frac{1}{4} kg $$
 * 2) $$ m = \frac{1 kg}{2} + \frac{1 kg}{4} $$
 * 3) $$ x = \frac{1}{2 K} + \frac{1}{4 K} $$
 * 1) $$ x = \frac{1}{2 K} + \frac{1}{4 K} $$
 * 1) $$ x = \frac{1}{2 K} + \frac{1}{4 K} $$