Ideal gases! One mole of ideal gas initially at 300 K is expanded from an initia
ID: 473887 • Letter: I
Question
Ideal gases! One mole of ideal gas initially at 300 K is expanded from an initial pressure of 10 atm to a final pressure of 1 atm. Calculate Delta U, q, w, Delta? and the final temperature T_2 for this expansion carried out according to each of the following paths. The heat capacity of an ideal gas is cv= 3R/2. An isothermal, reversible expansion an expansion against a constant external pressure of 1 atm in an insulated (adiabatic) system which prevents the gain or loss of heat. A reversible adiabatic expansion. An expansion against zero external pressure (into a vacuum) in an adiabatic system.Explanation / Answer
If V2 and P2 represent the conditions after expansion and V1 and P1 represent the conditions before expansion and at constant temperature P1V1=P2V2
W=-RT ln (V2/V1)= -RT ln (P1/P2)= -8.314*300*ln (10)= =5743 Joules ( work is –ve since it is done by the system and anything leaving is –ve)
From first law of thermodynamics
U= q+W. U=0 and q= -W= 5743 joules
b) An expansion against constant external pressure of 1 atm q=0 (since the process is adiabatic)
P2= 1atm and P1= 10 atm
Volume at 10 atm V1= nRT/P= 1*0.08206*300/10=2.4618 L
Underadiabatic conditions, ¥= Cp/CV
CV= 3/2 R =1.5R and CP= R+3/2R= 5/2 R =2.5R, ¥=5/3
T2 ( final temperature)= T1*{CV+R*P2/P1)/Cp}= T1*{( 1.5R+0.1R)/2.5R}=300*1.6/2.5=192 K
At T2=192K and P2= 1 atm
V2= 1*0.08206*192/1 =15.76 L
Work done against a constant pressure of 1 atm, W =-PdV =1* (15.76-2.4618) atm.L=-13.3Latm=-13.3*101.325 joules=-1347.6 Joules
U= q+W , since q=0 U= 0-1347.6 joules
H =n*Cp* (T2-T1)= 1* 2.5R*(192-300)=-2245 joules
c)
Reversible adiabatic expansion
q=0
T2= 192 K
W= R*(T2-T1)/( ¥-1)= {8.314*(192-300)/(5/3-1)}=-1360.47joules
U=q+W= -1360.47 joules
H =n*Cp* (T2-T1)= 1* 2.5R*(192-300)=-2245 joules
d) Work done against vaccum in an adiabatic system .
W = 0, q=0, U=0 and H=0
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