Question #53733
1 Answer
Explanation:
The idea here is that pressure and volume have an inverse relationship when temperature and number of moles of gas are being kept constant, as described by Boyle's Law.
When two parameters have an inverse relationship, increasing the value of one will cause the value of the other to decrease. Similarly, decreasing the value of one will cause the value of the other to increase.
You thus have
color(blue)(uarr) color(white)(.)"one parameter " stackrel(color(white)(color(red)("inverse relationship")aaa))(->) " the other"color(white)(.)color(blue)(darr)↑⏐⏐.one parameter inverse relationshipaaa−−−−−−−−−−−−−→ the other.⏐⏐↓
and
color(blue)(darr) color(white)(.)"one parameter " stackrel(color(white)(color(red)("inverse relationship")aaa))(->) " the other"color(white)(.)color(blue)(uarr)⏐⏐↓.one parameter inverse relationshipaaa−−−−−−−−−−−−−→ the other.↑⏐⏐
In your case, the pressure of the gas is decreasing from
Mathematically, Boyle's Law can be written as
color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))
Here
P_1 ,V_1 are the pressure and the volume of the gas at an initial stateP_2 ,V_2 are the pressure and the volume of the gas at a final state
You want to find the volume of the gas after the pressure is decreased, so rearrange to solve for
P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1
Plug in your values to find
V_2 = (12.7 color(red)(cancel(color(black)("kPa"))))/(8.4color(red)(cancel(color(black)("kPa")))) * "40.0 L" = color(darkgreen)(ul(color(black)("60. L")))
The answer must be rounded to two sig figs because you only have two sig figs given for the second pressure of the gas.
As predicted, the volume of the gas increased as a result of a decrease in pressure.
