A mixture of neon and oxygen gases, in a 9.77 L flask at 65 °C, contains 2.84 grams of neon and 7.67 grams of oxygen. The partial pressure of oxygen in the flask is ? atm and the total pressure in the flask is atm?

1 Answer
Nov 15, 2017

Answer:

#p_text(O₂) = "0.681 atm"; p_text(tot) = "1.081 atm"#

Explanation:

We can use the Ideal Gas Law to calculate the partial pressures.

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

or

#p = (nRT)/V#

The number of moles #n# is given by

#n = m/M#

So,

#p = (mRT)/(MV)#

Calculate #p_text(O₂)#

#m color(white)(l)= "7.67 g"#
#Rcolor(white)(l) = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#Tcolor(white)(l) = "65 °C" = "338.15 K"#
#M = "32.00 g·mol"^"-1""#
#V color(white)(l)= "9.77 L"#

#p = (7.67 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"color(red)(cancel(color(black)("·K"^"-1""mol"^"-1"))) × 338.15 color(red)(cancel(color(black)("K"))))/(32.00 color(red)(cancel(color(black)("g·mol"^"-1"))) × 9.77 color(red)(cancel(color(black)("L")))) = "0.681 atm"#

Calculate #p_text(Ne)#

#m color(white)(l)= "2.84 g"#
#Rcolor(white)(l) = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#Tcolor(white)(l) = "65 °C" = "338.15 K"#
#M = "20.18 g·mol"^"-1""#
#V color(white)(l)= "9.77 L"#

#p = (2.84 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"color(red)(cancel(color(black)("·K"^"-1""mol"^"-1"))) × 338.15 color(red)(cancel(color(black)("K"))))/(20.18 color(red)(cancel(color(black)("g·mol"^"-1"))) × 9.77 color(red)(cancel(color(black)("L")))) = "0.3997 atm"#

Calculate the total pressure

#p_text(tot) = p_text(O₂) + p_text(Ne) = "(0.681 + 0.3997) atm" = "1.081 atm"#