/t-1.38x 10-23 J/K h = 6.626x 10-34 Js, me-9.1 1x10 1. Below is a list of five o
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/t-1.38x 10-23 J/K h = 6.626x 10-34 Js, me-9.1 1x10 1. Below is a list of five of the most scien their contributions to physical chemistry th (1) Josiah Willard Gibbs (February 11, 1839 - April 28, 1903 (2) Ludwig Eduard Boltzmann (February 20, 1844 (3) Max Karl Ernst Ludwig Planck (April 23, 1 (4) Svante August (February mportant scientists. Describe the significance of field. (10%) - September 5, 1906) (4) Svante Arthenius (February 19, 1859 -October 2, 1927) enry Eyring (February 20, 1901 - December 26, 1981) itial temperature otz el monatomic gas at an initial pressure of 1.00 atm and pressure of 0.395 atm until equilibriun is reached (a) nsider 2 mole of an ideal temperature of 273.15 K. Assume it expandsadiabatically against a co ure? (b) What is the final volume? (c) How much work is done on the gas in this process? ) What is AU of the process? (e) What is H of the process? (20%) 3. What is the temperature of the flame of a Bunsen burner fed with a mixture of ingen and air? In the flame the detonating gas reaction: H2(g) + 0.5 02g akes place. Assume that a stoichiometric mixture is present above the reaction an that the reaction proceeds to completion. Air is composed of 4:1 mixture of N2 ar uhe that (HYH20s)--241.8 kJ/mol at 298 K. Explain why the calculated arue i s higher than the observed temperature. (20%)Explanation / Answer
For an ideal monoatomic gas at standard pressure and temperature, the ratio of specific heat capacity at constant pressure and constant volume = Cp/Cv
With three degrees of freedom at STP.
g=1+(2/f), let g be the ratio =Cp/Cv,f be the degree of freedom.
g=1+(2/3)= 5/3 ~ 1.6666
g= 1.67
Now, for adiabatic process:
Tg/P (g-1)= constant
Therefore, ( T1)g/ (P1)g-1= (T2)g/(P2)g-1
T1=273.15 K ; T2=?
P1=1 atm ; P2=0.395 atm
(T2)g= (T1)g (P2/P1)g-1
T2= T1 (P2/P1)1-(1/g)
1-(1/g)= 1- (3/5)= 2/5= 0.4
(P2/P1)0.4= (0.395/1)0.4= 0.68966
T2=( 0.68966)×273.15= 188.38 K
T1 (V1)g-1= T2(V2)g-1
v2=V1 (T1/T2)1/(g-1)
V2=2mole (273.15/188.38)3/2= 2×1.74= 3.9 mole
W= (P1V1-P2V2)/(g-1)
= (1×2- 0.395×3.94)/(5/3-1)= (0.4437)/(2/3)= (0.4437)×(3/2)=0.6655 mole atm
As the amount of heat supplied is zero thus
0= dU+P (dV)
Or dU= nCvdT
g=Cp/Cv= 5/3
Cp-Cv= R (gas constant )
= 8.3J/K /mol
(5/3)Cv- Cv= 8.3
Cv= (8.3×3)/2= 12.45
dU= 2× 12.45× (273.15-188.38)=2110.73J
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