** TEMPERATURE**

From Kelvin (K) to Celsius (°C ,former centigrade) scale:

T_{(°C) }= T_{(K)} - 273.15

From Kelvin (K) to Fahrenheit (°F) scale:

T_{(°F) }= (T_{(K)}*9/5) - 459.67

From Fahrenheit (°F) to Rankine (°R) scale:

T_{(°R) }= T_{(°F) }+ 459.67

** ENERGY**

1 J (joule) = 1 N·1m = 10^{7} erg

1 cal = 4.184 J

1 kWh(kilowatt hour) = 3.6·10^{6} J

1 BTU(British Thermal Unit) = 1055 J

** VOLUME**

1 ft^{3} (cubic feet) = 28.317 L (liters)

1 gal(US)(gallon) = 3.785 L

1 barrel = 42 gal(US) = 158.97 L

** Volume/Pressure/Temperature for (ideal) gases**

(**P**ressure) x (**V**olume) = (**n**° of moles) x **R** x (**T**emperature)

In short : **PV=nRT**

Pressure in atm. (1 atm = 101325 Pa = 760 mmHg)

Volume in Litres (L)

R = Universal constant = 0.082

Temperature in K

** Weight / mol**

**W**=weight of a given sample (solid,liquid or gas) in gram

n =**n**° of moles

**M** = molecular weight of the substance (example H_{2}O = 16+1+1=18)

**n = W/M** (example : in 1.00 L water there are 1000 g and 1000/18=55.55 mol of water)

**1 mol = 6.022·10 ^{23} particles** (molecules or atoms) (example : in 1.00 L water there are 55.55 mol of water and hence 55.55 * 6.022·10

** ΔH = ΔU + PΔV ; ΔF = ΔG + PΔV**

As usual rule in a chemical reaction, the only variation in volume (ΔV) considered is the difference between volume of gas in reactants and volume of gas in products.

If we call **Δn **the mole variation (gas moles of products - gas moles of reactants) we can write, according to above relationship,

**PΔV = Δn R T**

**ΔH = ΔU + Δn R T**

**ΔF = ΔG + Δn R T**