Combined Gas Law Solved Examples
Example 1: A gas has a 6L beginning volume and a 3L ending volume. Determine the gas’s final pressure so that the final temperature is 200 K and the initial temperature is 273 K. The starting pressure is 25 K Pa.
The given parameters are as follows:
Pi = 25 KPa, Vi = 6L, Vf = 3L, Ti = 273K , Tf = 200K
As per the combined gas law,Pi Vi / Ti = Pf Vf / Tf
So, substituting in the formula: 25 × 6/273 = Pf × 3/200
Pf = 36.626 KPa
Hence, the final pressure of the gas is 36.626 KPa.
Example 2: Determine the volume of a gas given Vi = 3L, Ti = 300K, Tf = 250K, Pi = 35 kPa and Pf= 50 kPa
Solution:
Given Parameters are: Pi = 35 kPa , Vi = 3L , Ti = 300K , Pf = 50 kPa , Tf = 250K
According to given parameters, we have an equation: Pi Vi / Ti = Pf Vf / Tf
Substituting values in the above equation: 35 × 3 / 300 = 50 × Vf / 250
Therefore, Vf = 1.75 L
Example: 2L of a gas at 350C and 0.833atm is brought to standard temperature and pressure (STP). What will be the new gas volume?
Solution:
P1 = 0.833 atm, V1 = 2.00L, T1 = 35℃ = 308K
P2 = 1.00 atm, T2 = 0℃ = 273K, V2 =?
Rearranging the equation algebraically for V2,
V2 = P1×V1×T2 / P2×T1
Now substitute the known quantities into the equation and solve.
V2 = 0.833 atm × 2.00L × 273K/1.00atm×308K
V2 = 1.48L
Combined Gas Law Formula
Combined Gas Law as the name suggest is combination of different gas laws. The different gas laws that combine to form Combined Gas Law include Boyle’s law, Charles’ Law and Gay Lussac’s law.
In this article, we will learn about combined gas law, its formula and derivation. We will also learn briefly the three gas laws individually.