Question #62744

2 Answers
Jul 24, 2017

Answer:

The sample contains 0.80 mol of #"N"_2#.

Explanation:

We can use the Ideal Gas Law to solve this problem:.

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

We can rearrange this equation to get

#n = (pV)/(RT)#

In this problem,

#p = "5.0 atm"#
#V = "4.0 dm"^3#
#R = "0.082 06 atm·dm"^3"K"^"-1""mol"^"-1"#
#T = "(32 + 273.15) K" = "305.15 K"#

#n = (5.0 color(red)(cancel(color(black)("atm"))) × 4.0 color(red)(cancel(color(black)("dm"^3))))/("0.082 06" color(red)(cancel(color(black)("atm·dm"^3"K"^"-1")))"mol"^"-1" × 305.15 color(red)(cancel(color(black)("K")))) = "0.80 mol"#

Jul 25, 2017

Answer:

There are 0.80 mol of nitrogen.

Explanation:

"But sir, what if the pressure unit is in pascals and the volume is in cubic metres instead of cubic decimetres?"

If

# p = 5.1 × 10^5color(white)(l) "Pa"#
#V = "0.0040 m"^3#
#R = "8.314 Pa·m"^3"K"^"-1""mol"^"-1"#
#T = "305.15 K"#

Then

#n = (5.1 × 10^5 color(red)(cancel(color(black)("Pa"))) × 0.0040 color(red)(cancel(color(black)("m"^3))))/( 8.314 color(red)(cancel(color(black)("Pa·m"^3"K"^"-1""mol"^"-1"))) × 305.15 color(red)(cancel(color(black)("K")))) = "0.80 mol"#