[21385]

I do not know how to approach this problem because I am not sure how to use the ideal gas law here.

One  Spring day  the  atmospheric  temperature  and pressure  are  15.08C  and  101.23  kPa
respectively,  and the  air contains  2%  by  volume of  water  vapor.    The  next  day  the
temperature is still 15.08C, but the pressure has dropped to 100.47 kPa.  Given that the
average relative molecular weight of the gases in dry air is 28.94, the water vapor content of
the air is now:
A.  0% B.  1% C.  3% D.  4% E.  5%

Can anyone help? Thanks a lot

[Borek]

I am before my morning coffee so I can be missing something obvious, but...

Are you sure this question is worded exactly as you have posted it? Because it seems to me that there is no correct answer between A-E.

[21385]

The question is exactly as worded in the original. The answer choices are (I presume) to the closest integer percent.

[Borek]

For ideal gases volume % is not pressure dependent. It is more or less conclusion of Avogadro's law.

Perhaps they are asking for mass %? And 95% of the question is there just to confuse you?

[AWK]

Let's take 100 moles of 2% air and the same volumes of air in both days.
Then p1/p2 =100/n2
where n2 is number of moles in this volume  in the second day.
when you set x - moles of H2O in this volume in the second day
then
n2=m2/M2
m2=(n2-x)28.94 + 18.02x
M2=(1-x/n2)28.84 + 18.02x/n2
solve for x, calculate 100x/n2 and compare to possible results

For goodness sake use as much significant figures as possible in the intermediate calculations.