Question #f6a82
1 Answer
Here's what I got.
Explanation:
The idea here is that STP conditions are usually given to you as a pressure of
0^@"C" = 0^@"C" + 273.15 = "273.15 K"0∘C=0∘C+273.15=273.15 K
Under these specific conditions for pressure and temperature,
Now, room conditions are usually given to you as a pressure of
20^@"C" + 273.15 = "293.15 K"20∘C+273.15=293.15 K
our goal here is to figure out the volume occupied by
color(blue)(ul(color(black)(PV = nRT)))
Here
P is the pressure of the gasV is the volume it occupiesn is the number of moles of gas present in the sampleR is the universal gas constant, equal to0.0821("atm L")/("mol K") T is the absolute temperature of the gas
Rearrange the equation
V/n = (RT)/P
Plug in your values to get
V/n = (0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 293.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("atm"))))
V/n = "24.07 L mol"^(-1)
This means that
Now all you have to do is use the molar mass of nitrogen gas to determine how many moles you have in your sample
35 color(red)(cancel(color(black)("g"))) * "1 mole n"_2/(28.0134color(red)(cancel(color(black)("g")))) = "1.249 moles N"_2
and use the molar volume of a gas at room conditions to find the volume it occupies
1.249 color(red)(cancel(color(black)("moles N"_2))) * "24.07 L"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(darkgreen)(ul(color(black)("30. L")))
The answer is rounded to two sig figs, the number of sig figs you have for the mass of nitrogen as.
SIDE NOTE Plug in the values for pressure and temperature you have at STP into this equation
V/n = (RT)/P
You should end up with
V/n = "22.4256 mol L"^(-1) ~~ "22.4 mol L"^(-1)
This is why we can say that 1 mole of any ideal gas occupies