Section 11–3:The Ideal Gas Law

Coach Kelsoe

Chemistry

Pages 383–385

Section 11–3 Objectives

State the ideal gas law. Derive the ideal gas constant and discuss its

units. Using the ideal gas law, calculate pressure,

volume, temperature, or amount of gas when the other three quantities are known.

Using the ideal gas law, calculate the molar mass or density of a gas.

Reduce the ideal gas law to Boyle’s law, Charles’s law, and Avogadro’s law. Describe the conditions under which each applies.

The Ideal Gas Law

In Chapter 10 we talked about the three quantities needed to describe a gas sample: pressure, volume, and temperature.

We can further characterize a gas sample by using a fourth quantity – the number of moles of the gas sample.

The number of molecules or moles present will always affect at least one of the other three quantities.

The Ideal Gas Law

Gas pressure, volume, temperature, and the number of moles are all interrelated. There is a mathematical relationship that describes the behavior of a gas sample for any combination of these conditions.

The ideal gas law is the mathematical relationship among pressure, volume, temperature, and the number of moles of a gas.

The equation for the ideal gas law is PV = nRT.

The Ideal Gas Law

PV = nRT, where P = pressure (in atmospheres) V = volume (in liters) n = number of moles of gas R = the ideal gas constant T = temperature (in Kelvins)

Since all these values are proportional, you could actually derive the ideal gas law from Boyle’s law, Charles’s law, and Avogadro’s law.

The Ideal Gas Constant

In the equation representing the ideal gas law, the constant R is known as the ideal gas constant.

The value of R depends on the units chosen for pressure, volume, and temperature.

For our calculations, we will let R be 0.0821 L·atm/mol·K. This means our measurements for pressure must be in atm, our volume must be in L, and our temperature must be in K.

The Ideal Gas Constant

We can solve for R by doing R = PV/nT at STP: P = 1 atm V = 22.41410 L n = 1 mol T = 273.15 K

R = PV/nT = (1 atm)(22.41410 L)/(1 mol)(273.15 K) R = 0.08205784 L·atm/mol·K

The Ideal Gas Constant

There are other values for R when you use different units.

R = PV/nT at STP: P = 760 mm Hg V = 22.41410 L n = 1 mol T = 273.15 K

R = PV/nT = (760 mm Hg)(22.41410 L)/(1 mol)(273.15 K) R = 62.4 L·mm Hg/mol·K

The Ideal Gas Constant

There are other values for R when you use different units.

R = PV/nT at STP: P = 101.325 kPa V = 22.41410 L n = 1 mol T = 273.15 K

R = PV/nT = (101.325 kPa)(22.41410 L)/(1 mol)(273.15 K) R = 8.314 L·kPa/mol·K

Finding P, V, T, or n From the Ideal Gas Law

You can easily solve for pressure, volume, temperature, or number of moles if you have three of the other values.

The value for R is always the same, so you’ll never have to solve for it.

P = nRT/V V = nRT/P n = PV/RT T = PV/nR

Sample Problem 11–3

What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? Given: V = 10.0 L, n = 0.500 mol, T = 298 K Unknown: P of N2 in atmospheres

PV = nRT P = nRT/V P = (0.500 mol)(0.0821 L·atm/mol·K)(298 K) / 10.0 L P = 1.22 atm

Sample Problem 11–4

What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0°C and 0.974 atm pressure? Given: P = 0.974 atm, n = 0.250 mol, T = 293.15 K Unknown: V of O2 in liters

PV = nRT V = nRT/P V = (0.250 mol)(0.0821 L·atm/mol·K)(293.15 K) / 0.974

atm V = 6.17 L

Sample Problem 11–5

What mass of chlorine gas, Cl2, in grams, is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure? Given: P = 3.50 atm, V = 10.0 L, T = 300 K Unknown: mass of Cl2 in grams

We don’t have a formula for mass, but we can find it by finding the value of n.

PV = nRT n = PV/RT n = (3.50 atm)(10.0 L)/(0.0821 L·atm/mol·K)

(300K) n = 1.42 mol Cl2; 1.42 mol x 70.90 g/1 mol = 101 g

Finding Molar Mass from the Ideal Gas Law

You can solve for molar mass once you’ve found the number of moles, but scientists have derived another form of the ideal gas law to save us time.

Since the number of moles (n) is equal to mass (m) divided by molar mass (M), we can substitute m/M for the letter n.

We can use PVM = mRT, where P is pressure, V is volume, M is molar mass (from periodic table), m is the sample mass, R is the ideal gas constant, and T is temperature.

Sample Problem 11–6

At 28°C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas? Given: P = 0.974 atm, V = 1.00 L, T = 301 K, m = 5.16 g Unknown: M of gas in g/mol

PVM = mRT M = mRT/PV M = (5.16 g)(0.0821 L·atm/mol·K)(301 K)/(0.974 atm)(1.00 L) M = 131 g/mol

Finding Density from the Ideal Gas Law

Density (D) is equal to mass (m) divided by (V), so therefore V = m/D. By subbing m/D for V, you can solve for density if you are given molar mass, and vice versa.

MP = DRT

Vocabulary

Ideal gas constant Ideal gas law