Here is an example of a Combined Gas Law problem.
Problem
You have a 50.00 L sample of an ideal gas at 28.46°C and 1.83 atm. The temperature increases to 51.69°C and the pressure decreases to 1.06 atm. The sample loses 15.0 % of its moles due to a leak in the container. What is the new volume of the sample?
Solution
First, make a list of all the variables.
We don't know the values of n_1n1 or n_2n2. Let's let n_1 = n " mol"n1=n mol. Then n_2 = 0.85n_1 = 0.85n " mol"n2=0.85n1=0.85n mol.
P_1 = "1.83 atm"P1=1.83 atm; V_1 = "50.00 L"V1=50.00 L; n_1 = n " mol"n1=n mol; T_1 = "(28.46 + 273.15) K" = "301.61 K"T1=(28.46 + 273.15) K=301.61 K
P_2 = 1.06 " atm"P2=1.06 atm; V_2 = "?"V2=?; n_2 = 0.85n " mol"n2=0.85n mol; T_ = "(51.69 + 273.15) K" = "324.84 K"T=(51.69 + 273.15) K=324.84 K
Insert these values into the Combined Gas Law.
(P_1V_1)/(n_1T_1) = (P_2V_2)/(n_2T_2)P1V1n1T1=P2V2n2T2
(1.83 cancel("atm") × 50.0" L")/(cancel(n" mol") × 301.61 cancel("K")) = (1.06 cancel("atm") × V_2)/(0.85 cancel(n " mol") × 324.84 cancel("K"))
"0.303 L" = 0.00383 V_2
Divide both sides of the equation by 0.00383.
V_2 = "0.303 L"/0.00383 = "79.0 L"
The new volume is 79.0 L.
