Ideal Gas Law at STP

Problem

Figure 1. Ideal gas law formula

The ideal gas law states:

$$PV = nRT$$

where

$P$ is the pressure measured in pascals (Pa).

$V$ is the volume measured in cubic meters (m³).

$n$ moles of a gas.

$R$ the gas constant 8.31 J/K/mol.

$T$ is the temperature measured in kelvins (K).

Part 1: Before 1982, STP (standard temperature and pressure) is defined as the condition when temperature is 0 degrees Celsius and the pressure is 1 atmosphere. Calculate the volume of 1 mole of any gas in STP in cubic meters and in liters.

Part 2: After 1982, STP (standard temperature and pressure) is defined as the condition when temperature is 0 degrees Celsius and the pressure is 1 bar. Calculate the volume of 1 mole of any gas in STP in cubic meters and in liters.

Solution

Part 1

Convert the pressure from atmospheres (atm) to pascals (Pa).

$$P = 1 \; atm = 101325 \; Pa$$

Convert the temperature from degrees Celsius (degs C) to kelvins (K).

$$T={0}^{\circ }C+273.15=273.15\;K$$

Use the ideal gas law to determine the volume.

$$PV = nRT$$

$$(101325\;Pa)V=(1\;mol)\left(8.31\;{\frac {J}{K\cdot mol}}\right)(273.15\;K)$$

$$V=0.0224\,m^{3}$$

$$V=22.4\,L$$

Part 2

Convert the pressure from bars (bar) to pascals (Pa).

$$P=1\;bar=100000\;Pa$$

Convert the temperature from degrees Celsius (degs C) to kelvins (K).

$$T={0}^{\circ }C+273.15=273.15\,K$$

Use the ideal gas law to determine the volume.

$$PV = nRT$$

$$(100000\;Pa)V=(1\;mol)\left(8.31\;{\frac {J}{K\cdot mol}}\right)(273.15\;K)$$

$$V=0.0227\;m^{3}$$

$$V=22.7\;L$$