Gas Laws, cont.

Combined Gas Law

We can combine Boyle's and Charles' relationships (T was part of the constant for Boyle's Law and P is part of the constant for Charles' Law) to give:

(PV)/T = constant.

V = an, where n = moles of stuff. So we have a linear relation between volume and moles; and V1/n1 = V2/n2

Ideal Gas Law

Ideal Gas Law ("Perfect Gas Law"): The constant for the combined law includes amount of stuff, and breaking that out we then get

(PV)/T = nR, or PV = nRT

where R = the gas constant with units appropriate to the various measurements. We will use atm, L, K, and moles, so that

R = 0.0821 L*atm/mole*K

I will base all of my examples on this equation because that requires a minimum of memorization. However you may find it easier to memorize a series of equations such as the "combined gas law equation" etc.

As an example, let's find the density of sulfur hexafluoride at a temperature of 25 °C and a pressure of 767 mmHg. The first thing we should do is determine what we want to know, mainly the mass of gas in a one Liter sample.

To do this we need to find 1) moles/L and 2) grams/mole. So let's write out the Ideal Gas Law:

PV = nRT

Rearranging to find moles/L,

n/V = P/RT

Next let's list what we know in this equation:

• P = 767 mmHg
• V = 1.000 L (assume one liter since want density in mol/L)
• n = ?
• R = 0.0821 L*atm/mole*K
• T = 25°C

Looking at these values we see that we need to convert P to atm and T to K in order to match the units in R!

• P = (767 mmHg)(1 atm/760 mmHg) = 1.009 atm
• T = 25 + 273.15 K = 298.15 K

Plugging these values into the gas law equation

n/V = P/RT = (1.009atm)/(0.0821L*atm/mole*K)(298.15K)

n/L = 4.122 x10-2 moles/one liter

The MW of SF6 = [32.07 + 6(19.00)] g/mol = 146.07 g/mol. Multiplying mol/L by g/mol then gives the density, D:

D = (4.122 x10-2 mol/L)(146.07 g/mol) = 6.0210 g/L = 6.02 g/L

Example: In an experiment the pressure of nitrogen inside a sealed, rigid container is found to be 15.6 atm at 557 °C. What was the pressure in the container when it was sealed in the lab at 25°C?

First look at the problem to see what changes and what doesn't.

• R is a constant so doesn't change.
• If the can is sealed, then n is constant.
• If it doesn't expand (its rigid), then V is constant.

Next write out the gas law equation,

PV = nRT

Rearranging to put the variables on one side:

P/T = nR/V, and since nR/V is constant, Po/To = Pf/Tf

Rearranging: Pf = (Po)(Tf/To)

Next, convert temperatures into Kelvins

• T1 = 25 + 273.15 = 298.15 K
• T2 = 557 + 273.15 = 830.15 K

Pf = (15.6 atm)(298.15 K)/(830.15 K)

= 5.133 atm = 5.13 atm.

Example: In an experiment the pressure of nitrogen inside a sealed, rigid container is found to be 5.133 atm at 25 °C. How many moles of gas are in the container if it has a volume of 500.0 mL?

• P = 5.133 atm
• V = 500.0 mL/1000 mL/L = 0.5000 L
• n = ?
• R = 0.0821 L*atm/mol*K
• T = 25° + 273.15° = 298.15 K

Rearranging, n = PV/RT

n = (5.133 atm)(0.5000 L)/(0.0821 L*atm/mol*K)(298.15 K)

n = 1.0485 x 10-1

n = 0.105 mol

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