Question

Question #71b77

Answer 1

`"150 mL"`

Explanation:

You actually have a few different ways of approaching this problem.

One way would be to use the ideal gas law equation to find how many moles of nitrogen gas you have in the sample, then use the molar volume of a gas at STP to find the sample's new volume.

A more direct way would be to use the combined gas law equation, since you know that the number of moles of gas remains constant.

If you start with the ideal gas law equation, you can write

`P_1V_1 = n * R * T_1 ->` for the initial conditions of the gas

and

`P_2V_2 = n* R * T_2 ->` for the final conditions of the gas

If you divide these two equations, you can get rid of the number of moles and of the unviersal gas constant, `R`

`(P_1V_1)/(P_2V_2) = (color(red)(cancel(color(black)(n * R))) * T_1)/(color(red)(cancel(color(black)(n * R))) * T_2) <=> (P_1V_1)/(P_2V_2) = T_1/T_2`

You can rearrange this equation to get the combined gas law form

`color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)`

You need to find the value of `V_2`, so you can say that

`V_2 = P_1/P_2 * T_2/T_1 * V_1`

Plug in your values and solve for `V_2` - do not forget to convert the temperature from degrees Celsius to Kelvin!

`V_2 = (650color(red)(cancel(color(black)("torr"))))/(760color(red)(cancel(color(black)("torr")))) * ((273.15 + 0)color(red)(cancel(color(black)("K"))))/((273.15 + 45.0)color(red)(cancel(color(black)("K")))) * "200 mL"`

`V_2 = "146.86 mL"`

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the initial volume of the gas

`V_2 = color(green)("150 mL")`

SIDE NOTE I highly recommend using the first approach as a way to double-check the result.

Use the ideal gas law equation to find the number of moles of nitrogen, then use the fact that at `0^@"C"` and `"1 atm"` one mole of any idel gas occupies exactly `"22.4 L"`.