Please help! I have been working on this homework assignment and I only have the
ID: 856597 • Letter: P
Question
Please help! I have been working on this homework assignment and I only have these last few questions to go! Thank you!
1) For Nitrogen Gas: a = 139x10^-3 m^3J/mol^2 and b = 39.1 mL = 39.1 x 10 ^-6 m^3
let n = 1 mol, T = 300 K and V = 22.4 L
a) Let b = 0, by what percent does the pressure change?
b) Let a = 0, by what percent does the pressure change?
c) If b = 39.1 mL in volume of a mole of N2 molecules, what is the approximate dimension of a Nitrogen atom in Amstrongs? Is this dimension consistant with the knowon dimension of N2?
2. If E = CvT - (a*n^2)/V show:
a) dE/dT holding V constant = Cv
b) dE/dV holding T constant = (a*n^2)/(V^2)
3. For N2 where Cv = 5/2R and T = 300K, n = 1, V = 22.4 L, and a = 139 x 10^-3 m^3*J/mol^2
What is the ratio of the CvT term (kinetic energy) to the attractive term (a*n^2)/V ?
5. Find the Temperature below which N2 will be cooled by expansion to a lower pressure
6. If T = 300 K , P1 = 20 atm and P2 = 1 atm
Find the temperature change (T2 - T1) for Nitrogen
Explanation / Answer
converting a into atmL2/mol2 hence we have to change the dimension.
1m3/mol2=9.869 atmL2/mol2
hence a=139x10^-3*9.869=1.37atmL2/mol2
now using the vander waals equation.
(P+an2/v2)(v-nb)=nRT
r=0.0821
n=1mol
T=300k
v=22.4L
and b also into litres=0.0391
putting in the equation we get
(P+1.37/(22.4)2)(22.4-0.0391)=0.0821*300
solving we get
P=1.0987atm
a)when a=0 we get Pnew(22.4-0.0391)=0.0821*300
we get Pnew=1.101476atm
hence % change in P would be (|(P-Pnew)|/P)*100
((1.101476-1.0987)/1.0987)*100=0.2526%
b)when b=0
we get
(Pnew+1.37/(22.4)2)22.4=0.0821*300
solving we get
Pnew=1.096821
hence % change in P=
(|(1.096821-1.0987)|/1.0987)*100=0.171%
c)b=4*NA*(volume of 1 atom of n2)
(volume of 1 atom of n2)=b/(4*NA) 0.0391/(4*6.022*1023)=1.6229*10-26 L
Yes this dimension is consistant with the knowon dimension of N2 as it approximately equals it=1.63 *10-26L
2)differentiating we get
a)(dE/dT)=Cv + an2/v2(dv/dT) since dv/dT=0 hence only term is Cv
b)(dE/dv)=Cv(dT/dv) + an2/v2 since dT/dv=0 hence only term is an2/v2
3) ratio would be (5/2RT)/(an2/v)=(2.5*0.0821*300)/(1.37/22.4)=1006.7737
5)63K which equals -210oC
6)P1/T1=P2/T2
20/300=1/T2
hence T2=15K
T2-T1=285K