Question

How many moles of gas are contained in 1.25 L at 303 K and 1.10 atm?

Answer 1

`"0.0553 moles"`

Explanation:

Your tool of choice here will be the ideal gas law equation, which looks like this

`color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "`

Here you have

`P` - the pressure of the gas
`V` - the volume it occupies
`n` - the number of moles of gas
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas

Now, take a look at the units used in the expression of the universal gas constant. The units you have for volume, temperature, and pressure must match those used in the expression of `R`.

`color(white)(aaaaaaaaaaacolor(Red)("Need")aaaaaaaaaaaaaaacolor(blue)("Have")aaaaa)`
`color(white)(aaaaa)color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaa)`
`color(white)(aaaaaaacolor(black)("Liters, L")aaaaaaaaaaaaacolor(black)("Liters, L")aaaaaaaaaaacolor(green)(sqrt())`
`color(white)(aaaaaaacolor(black)("Kelvin, K")aaaaaaaaaaaacolor(black)("Kelvin, K")aaaaaaaaaacolor(green)(sqrt())`
`color(white)(aaaaaaacolor(black)("Atmospheres, atm")aaaaacolor(black)("Atmospheres, atm")aaaacolor(green)(sqrt())`

Rearrange the ideal gas law equation to solve for `n`

`PV = nRT implies n = (PV)/(RT)`

Plug in your values to find

`n = (1.10 color(red)(cancel(color(black)("atm"))) * 1.25color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 303color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0553 moles")color(white)(a/a)|)))`

The answer is rounded to three sig figs.