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# 11.5: Universal Gas Law

Difficulty Level: At Grade Created by: CK-12

Compressed gases provide vital fuels for industry and for homes and farms in rural areas.

### Universal Gas Law

The combined gas law, $PV \propto T$ , is true for a particular sample of gas. If any gas is added or allowed to leak out, however, the relationship is lost. In order to get a relationship that is true for any sample of gas, it is necessary to incorporate a term for the amount of gas. From observations as simple as blowing up a balloon, it is clear that increasing the amount of gas increases the volume.

Because different gases have different weights per molecule, including a term for mass of gas does not produce a consistent equation. If, however, we include a term expressing the number of moles of gas rather than its mass, we can produce a constant proportionality. A mole is a unit representing the number of atoms present. The letter  $n$ is used to represent the moles of substance. Incorporating  $n$ into the equation yields  $PV \propto nT$ . If we insert a letter, $R$ , to represent the constant of proportionality, we get the normal form of the universal gas law , $PV = nRT$ .

The unit term for  $n$ is always moles and   $T$ is always in Kelvin. The units for pressure and volume, however, may vary. The value of  $R$ depends on the units that are used for pressure and volume.

 Pressure Units Volume Units Units for  $\underline{n}$ Units for  $\underline{T}$ Value of  $\underline{R}$ atm liters moles Kelvin 0.0821 L•atm/mol•K atm milliliters moles Kelvin 82.1 mL•atm/mol•K

Since the product of (liters)(atm) can be converted to joules, we also have a value for  $R$ where liters × atm have been converted to joules, $R = 8.314 \ \text{J/mol} \cdot \text{K}$ .  The two common values of the universal gas law $R$ constant are  $0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K}$ and $R = 8.314 \ \text{J/mol} \cdot \text{K}$ .

Most universal gas law problems are calculated at STP. STP stands for standard temperature and pressure,  which is the most commonly calculated temperature and pressure value. STP is defined as 1.00 atm and 0°C, or 273 K.

Example Problem: Determine the volume of 1.00 mol of any gas at STP.

Solution:  First isolate V from PV=nRT. Then plug in known values and solve.

$V=\frac{nRT}{P}=\frac{(1.00 \ \text{mol})(0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K})(273 \ \text{K})}{(1.00 \ \text{atm})}=22.4 \ \text{liters}$

For any gas at STP, one mole has a volume of 22.4 liters. This can be an extremely convenient conversion factor.

Example Problem: A sample of oxygen gas occupies 10.0 liters at STP.  How many moles of oxygen are in the container?

Solution:

$n=\frac{PV}{RT}=\frac{(1.00 \ \text{atm})(10.0 \ \text{L})}{(0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K})(273 \ \text{K})}=0.446 \ \text{moles}$

#### Summary

• The universal gas law is $PV = nRT$ , where  $P$ is pressure,  $V$ is volume,  $n$ is number of moles,  $R$ is the universal gas law constant, and  $T$ is the absolute temperature.
• The value of  $R$ varies depending on the units used for  $P$ and $V$ . Two common values are  $0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K}$ and $R = 8.314 \ \text{J/mol} \cdot \text{K}$ .
• STP is standard temperature and pressure; 273 K and 1.00 atm.
• One mole of a gas at STP has a volume of 22.4 liters.

#### Practice

The following video discusses the constant $R$ . Use this resource to answer the questions that follow.

1. Why is it important to have values for R in kPa, atm, and mmHg?
2. Why do the units of R include pressure, temperature, volume, and moles?

Instruction and practice problems involving the universal gas law:

#### Review

1. The initial pressure in a helium gas cylinder is 30 atm.  After many balloons have been blown up, the pressure in the cylinder has decreased to 6 atm while the volume and temperature remain the same.  What fraction of the original amount of gas remains in the cylinder?
2. Calculate the volume of 8.88 mol of helium gas at 20.0°C and 1.19 atm pressure.

Mar 13, 2013