Problem 1 2 mol of argon (considered as a monoatomic ideal gas) at 300 K and 3 b

Question

Problem 1 2 mol of argon (considered as a monoatomic ideal gas) at 300 K and 3 bar is cycled through the following three reversible steps: (1) The gas is allowed to expand isothermally such that its volume triples (2) Subsequently, the gas is compressed adiabatically to the initial pressure. (3) Finally the gas is isobarically cooled to its initial state at 300 K and 3 bar. For each step compute: (a) the work done by the gas (b) the heat transferred to gas (c) the change in the internal energy of the gas (d) the change in the enthalpy of the gas. Show your work.

Explanation / Answer

During isothermal expansion of an ideal gas, change in internal energy ( deltaU)=0

Enthalpy change, deltaH=0, work done by gas on surroundings = -nRTln (V2/V1)

Where n=2 moles T=300K V2/V1= volume ratio= 3

Work done = -2*300*8.314*ln(3)=-5480 Joules

Heat transferred (Q) from 1st law of thermodynamics, deltaU= Q+W

Q= -W = 5480 Joules

Pressure at the end of isothermal expansion , P2/P1= V1/V2, P2= 3/3= 1 bar

2. during adiabatic expansion, Q= 0, Pressure and temperature during adiabatic compression is related as T3/T2 = (P3/P2)(Y-1)/Y, Y= ratio of specific heats = 1.66

T3= temperature at the end of compression process, T2= 300K, P3= 3 bar and P1= 1bar

T3/300 = (3/1)(1.66-1)/1.66 , T3= 300*1.55= 465K

Work done = nR*(T2-T1)/(Y-1)= 2*8.314*(465-300)/0.66 =4157 Jolules

For adiabatic process, deltaU= Q+W, deltaU= 4157 Joules

Enthalpy change= nCP*deltaT , Cp for Argon = 0.52 J/g.k molar mass of Argon = 40 g/mole

Cp for Argon =0.52*40 J/mole.K= 20.8

Enthalpy change =2*40*0.52*(465-300)= 6864Joules

3. Third step involves cooling at constant pressure to 300K and 300 bar.

Work done = nR*deltaT= 2*8.314*(-465+300) =-2744 joules

Q= nCp*deltaT= 2*20.8*(-465+300)=- 6864 joules= deltaH

Cv= CP-R

Change in internal energy, deltaU= nCv*deltaT= 2*(20.8-8.314)*(300-465)= -4157Joules

2. in the gas phas reaction of 2A----->2B+C, there is an increase in pressure as the reaction proceeds. Hence an increase in presure is an indication of reaction giviing rise to products

initila pressure = nRT/V = 0.4*0.0821*(20+273)/ 6 =1.6 atm

let x= moles of C formed

hence after decomposition

A= 0.4-2x, B= 2x and C=x

total moles = 0.4+x , P= 3.3 atm, V= 6L, t= 482Deg.F= (482-32)/1.8 deg.c =250 deg.c = 250+273= 523K

P= 3.3= (0.4+x)*0.0821*523/6

0.4+x=0.46, x= 0.06

hence when the reaction is taking placem A= 0.4-0.2*0.06= 0.28, B= 0.12, C= 0.06 moles

A converted (0.4-0.28)/0.4 =0.3

partial pressure = mole fraction* total pressure

mole fraction= moles/total moles : A = 0.28/0.46=0.61, B= 0.12/0.46= 0.261, C= 1-0.61-0.261=0.129

Partial pressure, A= 0.61*3.3 =2 atm, B= 0.261*3.3 = 0.86 atm and c= 3.3-2.013-0.86= 0.43 atm

at 800 deg.c (1472 deg.F, deg.C= (1472-32)/1.8), the dissociation is complete, there is no A. 0.4 mole of A gives 0.4 mole of B and 0.2 mole of C as per the reaction, total moles = 0.4+0.2= 0.6

Hence Pressure = nRT/V= 0.6*0.0821*(800+273)/6=8.8 atm

at 1000 deg,c Pressure = 0.6*0.0821*1273/6= 10.45 atm

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.