STP question?

A storage tank at STP contains 18.0 kg of nitrogen (N2).

(a) What is the volume of the tank?

(answer in m^3)

(b) What is the pressure if an additional 10.0 kg of nitrogen is added?

(answer in Pa)

3 Answers

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  • Dr W
    Lv 7
    4 years ago
    Favorite Answer

    *** question #1 ***

    assuming STP means T = 0°C and P = 1atm

    for those chemical engineering students out there..

    .. V = (18.0kg N2) x (1 kgmol N2 / 28.0kg N2) x (22.41 m³ / kgmol @STP) = 14.4m³

    for those chemists out there

    ... . .. .18.0kg N2.. 1000g.. .1 mol N2... .... .. 22.41L.. . ... . 1 m³

    .. V = ----- ---- --- x --- ---- x ---- ---- ---- x ---- ---- ---- ----- x ----- --- = 14.4m³

    ... .. .. .. . .1.. ... .. . .1kg... .28.02g N2... mol N2 @STP... 1000L

    *** Question #2 ***

    there are a variety of different ways to do this.. given the other answerers answers, I'll show two.

    this way.. from the ideal gas law

    .. P1V1 / (n1T1) = P2V2 / (n2T2)

    rearranging

    .. P2 = P1 x (V1 / V2) x (T2 / T1) x (n2 / n1)

    since V is constant.. V1 = V2 and V1/V2 drops out

    since T2 = T1.... T2/T1 drops out

    leaving

    .. P2 = P1 x (n2 / n1)

    since n = mass / molar mass... and since molar mass is the same for N2 regardless of how much N2 is present, we can rewrite that as

    .. P2 = P1 x (mass2 / mass1)

    solving

    .. P2 = 1atm x (28.0kg / 18.0kg) x (101325 Pa / atm) = 1.58x10^5 Pa

    alternately...

    .. PV = nRT

    .. ... ..(28.0kg x 1000g/kg x 1mol/28.0g) x (0.08206 Latm/molK) x (273.15K).. .101325 kPa

    .. P = ----- ----- ---- ---- ---- ---- --- ---- ---- ---- ----- ---- ---- ---- ---- ---- ---- ---- -- x --- ---- --- ----

    ... .. ... ... ... ... .. ... .. ... .. .. .. .. ... 14.4x10^3 L... ... ... ... .... .. ... ... ... .. .. .. .. . .. .1 atm

    .. P = 1.58x10^5 Pa

    ************

    ************

    ************

    Mukherjee... <sigh>... what can I say that I haven't said to you in the past.

    .. (A) try again

    .. (B) you don't understand the basics

    .. (C) your logic is non-sequitur

    .. (D) delete your answer because it will no doubt cause more harm than good.

    ???

    let's go with "D" this time

    you wrote

    .. "Standard pressure and temperature referred to for gases are

    .. 1 atmosphere and 273 K or 101.325 Pa and 273K"

    that is incorrect. this is correct

    .. 1 atm = 101325 Pa = 101.325 kPa

    catch that?

    .. 101.325 KILOpascals

    and you wrote

    .. "18 kg N2 IS : 18 kg / 28 kgmol ( molecular weight of N2 )

    .. . = 0.642857142 moles of N2"

    also incorrect

    .. (1) the molecular "weight" of N2 is 28 AMU

    .. (2) the MOLAR MASS of N2 is 28g / mole = 28kg / kg-mole

    .. (3) 18kg is NOT 0.642857142 "moles"... it's 0.642857142 "kg-moles"

    ... . . .(there is a difference between mol and kg-mol) !!!!!

    then you wrote

    .. "0.0224 m3 / 1 mole x 0.0642657 moles ="

    which was off by a factor of 10 from your previous mole calc.. you should have written

    .. "0.0224 m3 / 1 mole x 0.642857142 moles ="

    .. ===>> notice how you mucked up the numbers 'cause you're carrying

    ... .. . ... . all those extraneous sig figs? <<===

    continuing.. that led to you making the mistake

    .. V = 1.43955168 x 10^-3 m3.

    the correct answer is

    .. V = 14.4m3... with 3 sig figs

    moving right along to question #2.. .. you wrote

    .. "With the addition of 10 kg".. blah blah blah... "the number of moles of N2

    .. . is 28 kg N2 / 28 kg N2/mole = 1 mole of N2.

    AGAIN.. confusing kg-mole with mole.. that is INCORRECT. It should read

    .. "28kg N2 x (1 kg-mole / 28.02kg) = 1.00 kg-mole of N2"

    then you wrote

    .. P1V1 / T1 = P2V2 / T2

    .. T1 = T2 = 273.15K

    .. V1 = V2 = 1.43955168x10^-3 m3.... (which is NOT correct)

    then realized you had no where to go with that so you glanced at Brent's answer and wrote

    .. P2 = 101.325 Pa x (1 mole / 0.644442857142mol) = 157.2288355 Pa

    which is also WRONG.

    .. (a) P1 = 101325 Pa... not 101.325 Pa..... back to that "kilo" thing again.

    .. (b) that equation.. P2 = P1 x (n2 / n1) does NOT follow from P1V1/T1 = P2V2/T2

    .. (c) the final answer is NOT 157.2288355 Pa

    ... . ... the sig figs are WRONG

    .... . .. the units are WRONG.

    .. (d) you have a transcribing error. 0.644442857142 vs 0.642857142

    .. .. . .caused by your abuse of sig figs.

    overall..

    .. terrible attempt.

    .. nonsensical logic

    .. non-existent sig figs leading to errors transcribing numbers TWICE

    .. poor understanding of mole, kg-mole, molar mass, etc...

    .. and you resorted to "cheating" and STILL GAVE THE WRONG ANSWER

    SO..

    .. (D) from above.... (delete your answer.. not worth fixing it) .. is my message to you this time

    ********

    ********

    One last thing.. And yes I'm going to use Mukherjees answer to demonstrate WHY we use DIMENSIONAL ANALYSIS in chemistry. To show the types of errors we can avoid with DA.

    look at the CALCULATIONS behind my two answers

    question #1... this is entered.. all at once in a calculator.

    .... engineering way... 18.0 / 28.02 * 22.41=

    .... chemists way.. .. . .18.02 / 1000 / 28.02 x 22.41 x 1000 =

    question #2

    .... 28.0 / 18.0 x 101325 =

    .. or

    .... 28*1000 / 28.02 * 0.08206 * 273.15 / 14.4e3 * 101325 =

    vs the CALCULATIONS behind Mukherjees answer.

    question #1

    .. 18 / 28 = 0.642857142... .. <=== conceptual error

    .. 22.4 / 1000 = 0.0224

    .. 0.0224 / 0.0642657 = 1.43955168x10^-3... <=== transcribing error

    question #2

    .. P1 = 101.325 Pa... ...<=== conceptual error

    .. n1 = 0.642857142... .<=== conceptual error

    .. n2 = 28 / 28 = 1.... .. .<=== conceptual error

    .. P2 = 101.325 * 1 / 0.644442857142 = 157.2288355 Pa... .

    ... .. .. ...this has both transcribing and conceptual errors

    again.. LEARN and USE the technique called "dimensional analysis". It will eliminate all those transcribing errors AND give you a logical process for solving problems.

  • 4 years ago

    Standard pressure and temperature referred to for gases are 1 atmosphere and 273 K or 101.325 Pa and 273K .

    18 kg N2 IS : 18 kg / 28 kgmol ( molecular weight of N2 ) = 0.642857142 moles of N2.

    At STP, N2 gas of 1 mole would occupy 22.4 liters volume = 22.4 / 1000 liters / m3 = 0.0224 m3 , so 0.6428571moles of N2 would occupy : 0.0224 m3 / 1 mole x 0.0642657 moles =1.43955168 x 10^-3 m3.

    With the addition of 10 kg addition into a fixed volume of a storage tank(at the same temperature of 273 k , the number of moles of N2 is 28 kg N2 / 28 kg N2/mole = 1 mole of N2.

    Thus number of moles of N2 increase would change the final pressure of the storage tank which can be found from ideal gas law at two different conditions ; p1v1/ T1 = p2v2 / T2 , where p1 = 101.325 Pa , T1= T2 = 273K , v1 = v2 = 1.43955168x 10^-3 m3 , n1 = 0.642857142 and n2 = 1 , so P2 = 1O1.325 Pa x 1/ 0.644442857142 = 157.2288355 Pa

  • Brent
    Lv 6
    4 years ago

    The definition of STP has recently changed.

    Classic definition is 1 atm (101.325 kPa) and 0°C. The molar volume in this case is 22.4 L.

    The new definition is 1 bar (100 kPa) and 0°C. The molar volume in this case is 22.7 L.

    Step 1 - Figure out moles of N2.

    n = 18 000 g * (1 mol / 28.02 g) = 642.4 mol

    Step 2 - Apply molar volume to find volume.

    Vol = 642.4 mol * 22.4 L/mol * 1 m³ / 1000 L = 14.4 m³

    (b) P2/P1 = m2/m1

    P2 = 101.325 kPa * 28.0 kg / 18.0 kg = 158 kPa

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