Question

Question #35f89

Answer 1

`"5.10 L"`

Explanation:

You know that, in addition to the number of moles of gas present in the sample, the temperature of the gas is being kept constant.

This should automatically tell you that you're dealing with a situation in which the volume of the gas will depend exclusively on its pressure.

When this happens, you can use the equation for Boyle's Law to help you find the new volume of the gas

`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

What this tells you is that when temperature and number of moles of gas are kept constant, pressure and volume have an inverse relationship.

Simply put, when pressure decreases, volume increases, and when pressure increases, volume decreases.

![http://www.peoi.org/Courses/Coursessp/chemintro/contents/frame6c.html]()

In your case, the pressure is decreasing from `"0.997 atm"` to `"0.977 atm"`, so right from the start you can say that the volume of the gas must increase as a result of this drop in pressure.

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 = (0.997color(red)(cancel(color(black)("atm"))))/(0.977color(red)(cancel(color(black)("atm")))) * "5.00 L" = "5.1024 L"`

Rounded to three sig figs, the answer will be

`V_2 = color(green)(|bar(ul(color(white)(a/a)"5.10 L"color(white)(a/a)|)))`

As predicted, the decrease in pressure lead to an increase in volume.