Question

Question #1c8e8

Answer 1

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_1` or `n_2`. Let's let `n_1 = n " mol"`. Then `n_2 = 0.85n_1 = 0.85n " mol"`.

`P_1 = "1.83 atm"`; `V_1 = "50.00 L"`; `n_1 = n " mol"`; `T_1 = "(28.46 + 273.15) K" = "301.61 K"`

`P_2 = 1.06 " atm"`; `V_2 = "?"`; `n_2 = 0.85n " mol"`; `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)`

`(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.