Ideal Gas Law

The Ideal Gas Law (P V = n R T) allows us to calculate various quantities for gases.

P V = n R T

P = Pressure (kilopascals) {kPa}
V = Volume (liters) {L}
n = number of moles [amount of substance] (moles) {mol}
R = universal gas constant (8.314 ^{kPa · L}/ _{mol · K})
T = Temperature (Kelvin) {K}

Calculations

P V = n R T

• R is always constant
• Within a given problem, n will typically be constant
• For P, V, T, usually 2 will vary and 1 will be constant
• The rest is just math and simple algebra

Steps for solving these calculations

1. Write the equation
2. Plug in all known values
3. Solve for the unknown
1. Combine values on each side
2. Divide, if needed, to solve

Basic Calculations

Sample 1:

How much pressure is needed to compress 6 moles of N₂ gas into a 15 liter container if it is held at a temperature of 298 K?
P V = n R T
P (15 L) = (6 mol) (8.314 ^{kPa · L}/ _{mol · K}) (298 K)
P (15 L) = 14865.432
P = ^14865.432/₁₅ = 991.0288 kPa ≈ 991 kPa

Sample 2:

How many moles of argon gas can be pressurized to 32 kPa in a 1.3 L container at a temperature of 300 K? P V = n R T
(32 kPa) (1.3 L) = n (8.314 ^{kPa · L}/ _{mol · K}) (300 K)
41.6 = n 2494.2
n = ^41.6/_2494.2 = 0.016678695 mol ≈ 0.0167 mol

Basic Relationships

Some relationships in the ideal gas law: P V = n R T

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Variables being Tested	Location in the Equation	Basic Relationship
P and V  (Boyle’s Law)	SAME SIDE	INDIRECT
T and V (Charles’ Law)	OPPOSITE SIDE	DIRECT
P and T	OPPOSITE SIDE	DIRECT
n and P	OPPOSITE SIDE	DIRECT
n and R	INVOLVES 'R'	NO RELATIONSHIP
V and R	INVOLVES ‘R’	NO RELATIONSHIP
T and n	SAME SIDE	INDIRECT

Before-and-After Calculations

Before and after calculations when you have a situation that is changing. The easiest way to solve these is to:

Set up the ideal gas law BEFORE (1) over
the ideal gas law AFTER (2)

P₁ V₁ = n₁ R T₁
P₂ V₂ = n₂ R T₂

Then cancel out all constants for that problem. ‘R’ will always cancel out. Solve for the remaining unknown.

Sample 3:

0.5 liters of O₂ gas is held at a temperature of 235 K. What temperature would the gas need to be raised to to keep the pressure constant but increase the volume to 3.1 L?

BEFORE: T₁ = 235 K, V₁ = 0.5 L (all else constant)
AFTER: T₂ = unk , V₂ = 3.1 L (all else constant)
CONSTANTS: P, n, R

Write the equations

P₁ V₁ = n₁ R T₁
P₂ V₂ = n₂ R T₂

Substitute in the givens

P₁ (0.5 L) = n₁ R (235 K)
P₂ (3.1 L) = n₂ R T₂

Remove the constants

(0.5 L) = (235 K)
(3.1 L) = T₂

Solve algebraically (Here, cross-multiply, then divide)

(0.5 L) T₂ = (235 K) (3.1 L)
 
(0.5 L) T₂ = 728.5
 
T₂ = ^728.5/_{(0.5 L)} = 1457 K

Combined Gas Law

This is a simplified version of solving Before-and-After calculations where moles is always considered to be constant, thus it cancels out.

This formula is given to you in Reference Table T.

P₁V₁ = P₂V₂
 T₁     T₂

Example: You have 16 liters of H₂O at 6 kPa of pressure and a constant temperature of 298 K. If the pressure increases to 12 kPa, what is the new volume?

P₁V₁ = P₂V₂
 T₁     T₂

6 kPa 16 L< = 12 kPaV₂
  298 K           298 K

Because temperature is constant, it will cancel out.

6 kPa 16 L = 12 kPaV₂

Combine, then solve.

96 kPa = 12 kPaV₂

V₂ = 96 L = 8 L
         12

Example: You have 18 moles of CO₂ at 22 kPa of pressure and at 300 K. If volume is held constant and temperature increases to 600 K, what would be the new pressure?

P₁V₁ = P₂V₂
 T₁     T₂

22 kPaV₁ = P₂V₂
  300 K     600 K

Volume is held constant, so it cancels out.

22 kPa = P₂
 300 K  600 K

Cross-multiply

(600 K)(22 kPa) = (300 K)P₂

Simplify and Divide

P₂ = 13200 kPa = 44 kPa
          300

Example: A 20 L vat of N₂ gas is held at 9 kPa pressure and 350 K. It's transferred to a 45 L vat with 12 kPa pressure. At what temperature will this new vat be held?

P₁V₁ = P₂V₂
 T₁     T₂

9 kPa(20) = (12)(45)
  350 K        T₂

Simplify

180   = 540
350 K    T₂

Cross-multiply

T₂(180) = (350 K)(540)

Simplify and Divide

T₂ = 189000 K = 1050 K
          180

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