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..
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
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
#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
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
Hence, rewrite
#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"#