Calculating Volume at Standard Temperature and Pressure

The ideal gas law specifies that the volume occupied by a gas depends upon the amount of substance (gas) as well as temperature and pressure. Standard temperature and pressure — usually abbreviated by the acronym STP — are 0 degrees Celsius and 1 atmosphere of pressure. Parameters of gases important for many calculations in chemistry and physics are usually calculated at STP. An example would be to calculate the volume that 56 g of nitrogen gas occupies.

Step 1

Get familiar with the ideal gas law. It can be written as: V = nRT/P. “P” is pressure, “V” is volume, n is the number of moles of a gas, “R” is the molar gas constant and “T” is temperature.

Step 2

Record the molar gas constant “R”. R = 8.314472 J/mole x K. The gas constant is expressed in the International System of Units (SI) and, hence, other parameters in the ideal gas equation must be in SI units as well.

Step 3

Convert pressure from atmospheres (atm) to Pascals (Pa) — the SI units — by multiplying by 101,325. Convert from degree Celsius to Kelvins — the SI units for temperature — by adding 273.15. Substituting these conversion in the ideal gas law produces a value of RT/P that is 0.022414 cubic meters/mole at STP. Thus, at STP, the ideal gas law can be written V = 0.022414n.

Step 4

Divide the mass of the gas weight by its molar mass to calculate n — the number of moles. Nitrogen gas has a molar mass of 28 g/mole, so 56 g of the gas is equivalent to 2 moles.

Step 5

Multiply the coefficient 0.022414 by the number of moles to calculate the gas volume (in cubic meters) at the standard temperature and pressure. In our example, the volume of the nitrogen gas is 0.022414 x 2 = 0.044828 cubic meters or 44.828 liters.

TL;DR (Too Long; Didn’t Read)

Helium has a molar mass of 4 g/mole, so 1 gram of the gas produces a balloon with a volume of 5.6 liters — a little over a gallon — at STP. If you filled the balloon with 1 gram of nitrogen gas instead, the balloon would shrink to 1/7 of that size, or 0.81 liters.