Question #e5462

You are indeed dealing with a gas law problem, a Boyle's Law problem to be precise.

According to Boyle's Law, when temperature and number of moles of gas are kept constant, volume and pressure have an inverse relationship.

When pressure decreases, volume increases, and when pressure increases, volume decreases.

Mathematically, Boyle's Law is expressed as

#color(blue)(|bar(ul(color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#, where

#P_1#, #V_1# - the pressure and volume of the gas at an initial state
#P_2#, #V_2# - the pressure and volume of the gas at a final state

In your case, you know that a sample of gas occupies a volume of #"473 mL"# at a pressure of #"1380 torr"#.

Standard pressure is defined as a pressure of #"100 kPa"#, or #"0.98692 atm"#. You can convert this pressure to mmHg by using the conversion factor

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 torr")color(white)(a/a)|)))#

The final pressure of the gas will thus be

#0.98692color(red)(cancel(color(black)("atm"))) * "760 torr"/(1color(red)(cancel(color(black)("atm")))) = "750.06 torr"#

So, the pressure of the gas decreases from #"1380 torr"# to #"750.06 torr"#. This tells you that you can expect the volume of the gas to increase.

Rearrange the equation to solve for #V_2#

#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#

Plug in your values to get

#V_2 = (1380 color(red)(cancel(color(black)("torr"))))/(750.06color(red)(cancel(color(black)("torr")))) * "473 mL" = "870.25 mL"#

Rounded to three sig figs, the answer will be

#V_2 = color(green)(|bar(ul(color(white)(a/a)"870 mL"color(white)(a/a)|)))#