The cool thing about STP problems is the 1 mole of any ideal gas occupies specifically 22.7 L - this is well-known as the molar volume the a gas in ~ STP.

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STP problems imply a temperature that 273.15 K and a pressure of 100 kPa. When tose problems are met, 1 mole of any kind of ideal gas will have actually a volume the 22.7 L.

So, if 1 mole occupies a volume the 22.7 L, 2 moles will occupy a volume double as big.

#2cancel("moles") * "22.7 L"/(1cancel("mole")) = "45.4 L"#

Likewise, 0.5 moles will occupy half the volume 1 mole occupies.

#0.5cancel("moles") * "22.7 L"/(1cancel("mole")) = "11.35 L"#

SIDE NOTE Many digital sources and textbooks still list the old STP problems of 273.15 K and also 1 atm. Under this conditions, the molar volume the a gas is actually 22.4 L.

If your teacher or textbook still offers that value, simply redo the calculation using 22.4 L instead of 22.7 L.


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Truong-Son N.
Jun 30, 2015

I"m walking to i think the appropriate case, yet if you desire to recognize a an ext accurate value, I"ll display that below.

Using the Ideal Gas Law:

#PV = nRT#

At STP, #P = "1 bar"#, #R = "0.083145 L"*"bar""/mol"cdot"K"#, and also #T = "273.15 K"#. Therefore, with #n = "2 mol"#:

#V = (nRT)/P = <(2 cancel("mol"))(("0.083145 L"cdotcancel("bar"))/(cancel("mol"*"K")))(273.15 cancel("K"))>/(1 cancel("bar"))#

#=# #color(blue)("45.422 L")#

A much more accurate answer would certainly be #color(blue)("44.968 L")#, utilizing the van der Waals equation of state:

#P = (RT)/(barV - b) - a/(barV^2)#

In this case, we"d need to know the constants #a# and #b# because that #Cl_2# specifically.

#a = "6.343 bar/L"^2"mol"^2#

#b = "0.05422 L/mol"#

Now we should solve because that #barV#:

#P = (RTbarV^2 - a(barV - b))/((barV - b)barV^2)#(cross-multiply)

#PbarV^2(barV - b)= RTbarV^2 - abarV + ab#(subtract through, multiply by denominator)

#PbarV^3 - bPbarV^2= RTbarV^2 - abarV + ab#(distribute)

#PbarV^3 - bPbarV^2 - RTbarV^2 + abarV - abdominal = 0#(move things around)

#PbarV^3 - (bP+RT)barV^2 + abarV - abdominal muscle = 0#(factor)

And currently to deal with this, one means I know of is to use the Newton-Raphson approximation method to strategy the answer indigenous above.

Let:

#P = color(green)(A) = color(green)("1 bar")#

#bP + RT = color(green)(B) = ("0.05422 L/mol")("1 bar") + ("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K") = color(green)("22.76527675 L"cdot"bar/mol")#

#a = color(green)(C) = color(green)("6.343 bar/L"^2"mol"^2)#

#ab = color(green)(D) = "6.343 bar/L"^2"mol"^2 cdot "0.05422 L/mol" = color(green)("0.34391746 bar/L"cdot"mol"^3)#

#barV = X# in #"L/mol"#

Now we have:

#AX^3 - BX^2 + CX - D = 0#

What you deserve to do is use the complying with formula because that the Newton-Raphson method:

#color(darkblue)(X_"new" = X_"old" - (f(x))/(f"(x)))#

#f(x) = AX^3 - BX^2 + CX - D##f"(x) = 3AX^2 - 2BX + C#

Thus:

#color(darkred)(X_"new" = X_"old" - (AX^3 - BX^2 + CX - D)/(3AX^2 - 2BX + C))#

Now what you might want to perform is keep a number into each variable in this equation, because you"ll have to keep recalling a lengthy expression. The coefficients are given above.

For #X_"old"#, simply pick a number and guess. I would certainly pick one of two people something much less than #1#, something close to #1#, or something in between #2# and also #30#. This will offer you 3 answers.

Then, if you usage a TI calculator, compose this right into your calculator to store variables:

#2->X##1 ->A#etc.

and this to solve:

#(X - ((AX^3 - BX^2 + CX - D)/(3AX^2 - 2BX + C))) -> X#

and then press go into until your answer stop changing.

Using #X_"old" = 2#, I gained #X_"new" = 0.2078358407# after ~ 7 times pressing Enter. Understanding that, ns would try something under #0.207#.

Using #X_"old" = 0.1#, I obtained #X_"new" = 0.0735975286# after ~ 6 times pushing Enter. Having tried something near #1# and something listed below #1#, let"s try something high.

Using #X_"old" = 20#, I gained #X_"new" = 22.48384338# ~ 6 times pushing Enter.

This means then the the three answers I gained are:

#"0.2078358407 L/mol"#

#"22.48384338 L/mol"#

The shortest answer coincides to the volume of 1 mol of liquid chlorine. The highest answer is the volume the 1 mol of gaseous chlorine. The center answer is nonphysical, for this reason we"re not going to usage it. This watch something favor this:

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If you see the horizontal line listed below "Two-Phase", the is the intersection between the three answers. The liquid is ~ above the left, and the gas is top top the right.

Thus, to gain the more reality result for gaseous chlorine:

#"22.48384338 L/mol" * "2 mol" = 44.96768676 -> color(blue)("44.968 L")#

compared come the ideal situation of #color(blue)("45.422 L")#.

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Checking its compressibility factor:

#Z = (PbarV)/(RT) = ((1)(22.48384338))/((0.083145)(273.15)) = 0.989995#

Since #Z , #Cl_2# is easier to compress in real life than its appropriate version, so its volume in ~ STP should be smaller sized in genuine life than in the right case, which us got.