Calculate the internal energy of 2 moles of argon gas (assume ideal behavior) at

Question

Calculate the internal energy of 2 moles of argon gas (assume ideal behavior) at 298 K. Suggest two ways to increase its internal energy by 10 J. A 1.00 mole sample of ammonia at 14.2 atm and 25 degrees Celsius is contained in a cylinder with a movable piston that expands against a constant external pressure of 1.00 atm. The pressure and volume of the gas are 1.00 atm and 24.5 L at equilibrium. Calculate the final temperature of the sample. Calculate the values of q, w, and delta U. Molar heat capacity at constant pressure is 35.66 J K^-1 mol^-1.

Explanation / Answer

(1)

Argonn=2 mol

T=298 K

Internal energy U=translation energy=3/2 nRT=3/2*2*8.314*298=7432.716 J

Increasing enegy by 10 J

(1) increase no of moles upto 2.0026

(2) Heat up the system upto 298.4 K temperature.

(2) ammonia

n=1 mol

P1=14.2 atm

T1=25 0C=298.15 K

P2=1 atm

V2=24.5 l

Using ideal gas law

PV=nRT

R=0.082 lit.atm/molK

T2=P2V2/nR=1*24.5/0.082*1=298.78 K

Now for finding work

W= -PV

So first we have to find

V1=nRT1/P1=1*0.082*298.15/14.2=1.72 lit

So work W =-1*101325* Pa(24.5-1.72)*10-3 m3=-2308.18 J=-2.308 kJ

U=CvT

Cv=Cp-R=35.66-8.314=27.341 J K-1mol-1

U=27.341*(298.78-298.15)=17.22483 J

U=q+w

q=U-w=17.22483-(-2308.18)=2325.40 J

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