# Question #1c8e8

Apr 13, 2014

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 ${n}_{1} = n \text{ mol}$. Then ${n}_{2} = 0.85 {n}_{1} = 0.85 n \text{ mol}$.

${P}_{1} = \text{1.83 atm}$; ${V}_{1} = \text{50.00 L}$; ${n}_{1} = n \text{ mol}$; ${T}_{1} = \text{(28.46 + 273.15) K" = "301.61 K}$

${P}_{2} = 1.06 \text{ atm}$; ${V}_{2} = \text{?}$; ${n}_{2} = 0.85 n \text{ mol}$; ${T}_{=} \text{(51.69 + 273.15) K" = "324.84 K}$

Insert these values into the Combined Gas Law.

$\frac{{P}_{1} {V}_{1}}{{n}_{1} {T}_{1}} = \frac{{P}_{2} {V}_{2}}{{n}_{2} {T}_{2}}$

$\left(1.83 \cancel{\text{atm") × 50.0" L")/(cancel(n" mol") × 301.61 cancel("K")) = (1.06 cancel("atm") × V_2)/(0.85 cancel(n " mol") × 324.84 cancel("K}}\right)$

$\text{0.303 L} = 0.00383 {V}_{2}$

Divide both sides of the equation by 0.00383.

${V}_{2} = \text{0.303 L"/0.00383 = "79.0 L}$

The new volume is 79.0 L.