A container has a volume of #72 L# and holds #6 mol# of gas. If the container is compressed such that its new volume is #3 L#, how many moles of gas must be released to maintain a constant temperature and pressure?

Question

A container has a volume of #72   L# and holds #6   mol# of gas.  If the container is compressed such that its new volume is #3 L#, how many moles of gas must be released to maintain a constant temperature and pressure?

Emily Abbate · Accepted Answer

#0.25 "mol"# of gas must be released.

Applying Avogadro's law here: #v_1/n_1 = v_2/n_2# The initial conditions are denoted by the number 1, and the final conditions are represented by the number 2. • Determine the variables you know and don't know: #color(red)("Knowns:"# #v_1#= 72 L #v_2#= 3 L #n_1#= 6 mol #color(magenta) ("Unknowns:"# #n_2# • Rewrite the equation to find the total number of moles in the end: #n_2=(v_2xxn_1)/v_1# • Enter the provided values to determine the total number of moles: #n_2=(3cancelLxx6mol)/(72\cancel"L")# = # 0.25 mol#

Elena Albarran · Answer

undefined Using the ideal gas law, we can solve for the number of moles of gas released. According to Avogadro's law, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas present. So, we can set up the equation:

$ \frac{V_1}{n_1} = \frac{V_2}{n_2} $

Given that $V_1 = 72 \, \text{L}$, $n_1 = 6 \, \text{mol}$, and $V_2 = 3 \, \text{L}$, we can solve for $n_2$:

$ \frac{72}{6} = \frac{3}{n_2} $

$ n_2 = \frac{3 \times 6}{72} = \frac{1}{4} $

So, $ n_2 = 1.5 \, \text{mol} $