An oxygen tank kept at 20oC contains 28.0 moles of oxygen and the gauge reads 31.0 atm. After two weeks, the gauge reads 10.5 atm. How many moles of oxygen were used during the two-week period?

An oxygen tank kept at 20oC contains 28.0 moles of oxygen and the gauge reads 31.0 atm. After two weeks, the gauge reads 10.5 atm. How many moles of oxygen were used during the two-week period?

9.48 moles
20.5 moles
9.25 moles
7.5 moles
18.5 moles

I think its either 18.5 or 9.25..

1 Answer
Aug 9, 2018

n = 9.48 color(white)(l) "mol"

Explanation:

The Ideal Gas Law states that

P * V = n * R * T

Rearranging gives

n/P = V / (R* T)

... meaning that ratio between the pressure P an ideal gas exerts on a container of a definite volume V and temperature T to the number of moles of gas particles n present in this gas takes a constant value. That is:

n_1/P_1 = V / (R* T) = n_2/P_2

This question implies that the temperature when both measurements take place is constant at 20 color(white)(l) ""^"o""C". It explicitly states that the container (the gas tank) is closed and has a constant volume. Thus the relationship n_1//P_1 = n_2 // P_2 holds where n_1 and P_1 are measurements of the gas at the beginning of the two weeks and n_2 and P_2 at the end of the period.

The question supplies the following information:

  • n_1 = 28.0 color(white)(l) "mol"
  • P_1 = 31.0 color(white)(l) "atm"
  • P_2 = 10.5 color(white)(l) "atm"

... and is asking for the value of n_2.

Hence, rewrite n_1//P_1 = n_2 // P_2 to isolate n_2:

n_2 = n_1 * (P_2) / (P_1)
color(white)(n_2) = 28.0 color(white)(l) "mol" xx (31.0 color(white)(l) color(red)(cancel(color(black)("atm"))))/(10.5 color(white)(l) color(red)(cancel(color(black)("atm")))
color(white)(n_2) = 9.48 color(white)(l) "mol"