Combined Gas Law
You guessed it the combined gas law is a combination of Boyle’s and Charles’
The difference is that we have all three, pressure, temperature and volume in the equation.
You will be given five of six variables and have to determine the sixth…
PV
PV
1 1
 2 2
T1
T2
As with all the gas laws, temperature needs to be in degrees Kelvin. Pressures and
volumes just need to be in the same units as each other.
Example:
32 liters of a gas at 300 degrees kelvin and 200 kPa is compressed to 400 kPa and 15
liters, what is the new temperature?
-
Make sure pressures and volumes are in the same units, temperatures in
degrees Kelvin.
-
Substitute, into the equation
-
cross multiply, divide,
-
check sigfigs and units.
200 32
300
=
400 15
T2
Cross multiply and divide
T2 x 200 x 32=400 x 15 x 300
T2= (400 x 15 x 300)/(200 x 32)
T2=
281.25
280 Liters
Try the next one yourself:
We take 375 ml of H2 at 1.00 atm and 23 degrees Celsius, what will the new pressure be
if it expands to 1.50 liters at a new temperature of 30 degrees celsius?
First - Temperatures must be in Kelvin
T1= 23 C = 23 + 273 = 296 Kelvin
T2= 30 C = 30 + 273 = 303 Kelvin
Second – make all like units the same make 375 ml into liters, V1=0.375 L
Third – Substitute into Combined Gas Law
P1V1/T1=P2V2/T2
1.00 atm x 0.375 ml / 296 = P2 x 1.50 / 303
Cross Multiply
1.00 x 0.375 x 303 = P2 x 1.50 x 296
Divide
P2 = (1.00 x 0.375 x 303) /(1.50 x 296)
P2 = 113.6 / 444
P2 = 0.256 atm
Or …
Solve combined gas law for P2 at step 3
P1V1/T1=P2V2/T2
P2 = (P1 x V1 x T2)/(T1 x V2)
Then substitute.
Ideal Gas Law
PV = nRT
The “n” = number of moles, R is the ideal gas constant = 8.31 kPa L / (mol deg K)
As with the previous gas laws, temperature must be in degrees Kelvin.
The remaining quantities must be in the same units in the gas constant.
Pressure in kPa, Volume in Liters (or dm3), amount of matter , n, in mols.
You will be given four of the five quantities and be asked to solve for the fifth.
Make sure your variables are in units of the gas constant
Substitute and solve.
Example
How many moles of an ideal gas are contained in 35600 milliliters at 250 deg C at 2.27
atm?
Convert temp to K 250 C + 273 = 523 K
Convert pressure to kPa
2.27 atm x 101.3 kPa = 230 kPa
1.0 atm
Convert volume to liters
35600 ml x 1 Liter =35.6 liters
1000 ml
PV = nRT
230 kPa x 35.6 L = n x 8.31 kPa L /(mol deg K) x (523 deg K)
n= 230 x 35.6 / (8.31 x 523) moles
n= 1.88 mol