Question #8cf3a

Assuming this behaves as an idea gas, we have the equation:
#PV=nRT#, where:

• #P# = Pressure (#Pa#)
• #V# = Volume (#m^3#)
• #n# = Number of moles (#mol#)
• #R# = Gas constant (#~8.31Jcolor(white)(l)mol^(-1)color(white)(l)K^(-1)#)
• #T# = Temperature (#K#)

To find #T# we can rearrange to get #T=(PV)/(nR)#

#P=33MPa=33*10^6Pa#
#V=3000mm^3=3*10^(-6)m^3#
#n=m/M_r=1/(2*14)=1/28mol#
#R=8.31Jcolor(white)(l)mol^(-1)color(white)(l)K^(-1)#

#T_1=((33*10^6)(3*10^(-6)))/(1/28*8.31)=333.5740072K=334K#

#T_1~~334K#

To find the new pressure at #30144mm^3#, we can use: #P=(nRT)/V#

#P=(1/28*8.31*((33*10^6)(3*10^(-6)))/(1/28*8.31))/(3.0144*10^(-5))#

#=(cancel(1/28*8.31)*((33*10^6)(3*10^(-6)))/(cancel(1/28)*cancel(8.31)))/(3.0144*10^(-5))#

#=((33*10^6)(3*10^(-6)))/(3.0144*10^(-5))~~3.28*10^6Pa=3.28MPa#

#P_2~~3.28*10^6Pa=3.28MPa#

Now to compare this with Van der Waals equation of state:
#(P+(n^2a)/V^2)(V-nb)=nRT#:

The only new values are #a# and #b# where:

• #a# = the amount of attraction between particles (#Jcolor(white)(l)m^2color(white)(lmol^(-2)#)
• #b# = volume of a gas per mole (#m^3color(white)(l)mol^(-1)#)

#a# and #b# are constants for specific gases.

For #N_2#, #a=0.137Jcolor(white)(l)m^2color(white)(l)mol^(-2)# and #b=3.87*10^(-5)m^3color(white)(l)mol^(-1)#

To solve for #T# we use:
#T=((P+(n^2a)/V^2)(V-nb))/(nR)#

#=(((33*10^6)+0.137(1/(28(3*10^(-6)))^2))(3*10^(-6)-(3.87*10^(-5))/28))/((8.31/28))#

Which after calculating comes out as #T_1=285.73397349318~~286K#

Now to solve for the new pressure at a constant #T, n, R, a# and #b#

#(P_2+0.137/(28(3.0144*10^(-5)))^2)((3.0144*10^(-5))-(3.87*10^(-5))/28)=(285.73397349318*8.31*1/28)#

#P_2=(285.73397349318*8.31*1/28)/((3.0144*10^(-5))-(3.87*10^(-5))/28)-0.137/(28(3.0144*10^(-5)))^2#

#=312612767184732201015625/113425766518212918#

#=2756100.10654Pa~~2.76*10^6Pa#

I got #"27.563 bar"#, or #2.7563 xx 10^6 "Pa"#.

The form of the van der Waals (vdW) equation I know is:

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

where:

• #a# and #b# are the vdW constants.
• #P# is the pressure in #"bar"#.
• #barV# is the molar volume in #"L/mol"#.
• #T# is the temperature in #"K"#.
• #R = "0.083145 L"cdot"bar/mol"cdot"K"# is the universal gas constant.

You seem to have given:

• #P_1 = "33 MPa"#
• #V_1 = "3000 mm"^3#
• #T_1 = ? -= T#
  #" "#
• #P_2 = ?#
• #V_2 = "30144 mm"^3#
• #T_2 = T_1 -= T#

Van der Waals constants can be found here or in your text. I see:

#a = "1.370 bar"cdot"L"^2"/mol"^2#
#b = "0.0387 L/mol"#

Therefore, we can solve for the temperature that this gas is at:

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

#T = [P + a/(barV^2)][(barV - b)/R]#

Now, we'll need to convert our variables to the proper units.

Initial Variables

#P_1 = 33 cancel"MPa" xx (10^6 cancel"Pa")/(cancel"1 MPa") xx ("1 bar")/(10^5 cancel"Pa") = ul"330 bar"#

#V_1 = 3000 cancel("mm"^3) xx (cancel("1 m")/(1000 cancel"mm"))^3 xx (("10 dm")/(cancel"1 m"))^3#

#= "0.003 dm"^3 = "0.003 L"#

#cancel("1 g N"_2) xx "1 mol"/(28.014 cancel("g N"_2)) = "0.0357 mol N"_2#

#=> barV_1 = "0.003 L"/"0.0357 mols" = ul"0.08404 L/mol"#

Final Variables

#V_2 = 30144 cancel("mm"^3) xx (cancel("1 m")/(1000 cancel"mm"))^3 xx (("10 dm")/(cancel"1 m"))^3#

#= "0.030144 dm"^3 = "0.030144 L"#

#=> barV_2 = "0.030144 L"/"0.0357 mols" = ul"0.84445 L/mol"#

As a result,

#T = ["330 bar" + ("1.370 bar"cdot"L"^2"/mol"^2)/(("0.08404 L/mol")^2)][("0.08404 L/mol" - "0.0387 L/mol")/("0.083145 L"cdot"bar/mol"cdot"K")]#

#=# #ul"285.73 K"#

So, at constant temperature and the new molar volume of #"0.84445 L/mol"#, the new pressure needed is:

#color(blue)(P_2) = (RT)/(barV - b) - a/(barV^2)#

#= ("0.083145 L"cdot"bar/mol"cdot"K" cdot "285.73 K")/("0.84445 L/mol" - "0.0387 L/mol") - ("1.370 bar"cdot"L"^2"/mol"^2)/(("0.84445 L/mol")^2)#

#=# #color(blue)ul("27.563 bar")#