A closed, hydrostatic, ideal-gas system is caused to follow the following revers

Question

A closed, hydrostatic, ideal-gas system is caused to follow the following reversible cycle. To prevent round-off errors in the calculations, report all answers to four (4) significant figures. Report energy values for this problem in liter-aim (extensive). (a) Calculate reversible work (extensive) for each path (a, b, c, d). In each case you roust (a) begin with the expression delta w = PdV, (b) Use the ideal gas law to obtain a symbolic differential expression on the specifed path of the form w = integral delta w = (constants) integral f(V)dV, (c) Integrate between defined end states to obtain the integrated symbolic equation, (d) Substiture values to obtain the final numerical result. Stale whether work is done on the system or by the system. (b) Calculate Delta U(extensive) for each path (a, b, c, d). In this case, you may use the fact that Delta U is independent of path and for each path use any convenient path to calculate Delta U. (c) Summarize in a neat table the results for w and Delta U for each path segment and for die entire cyclic process. The molar heat capacity at constant volume is c_v = (3/2)R energy/(mole K) R = 0.082 liter-atm/(mole K) 1 atm = 1.013 times 10^5 Pa r = 8.314 J/(mole K) 1 liter-atm = 101.3 joules 1000 liters = 1 m^3

Explanation / Answer

given

P1 = 1 atm = 1.01*10^5 Pa

T1 = 300 K

V1 = 6 lit = 0.006 m^3

using ideal gas law

PV = nRT

n = P1V1/RT1 = 1.01*10^5*0.006/8.31*300 = 0.243

process 1 -> P/V = constant = k1

P2 = 2atm

T2 = ?

V2 = ?

using P1/V1 = P2/V2

V2 = P2*V1/P1 = 2*0.006/1 = 0.012 m^3

using ideal gas law

P2V2 = nRT2

2*1.01*10^5 * 0.012 = 0.243*8.31*T2

T2 = 1200.398 K

work done in process 1, dW = PdV

now, P = k1*V

so, dW = k1*VdV

integrating

work done by gas, W1 = P1*(V2^2 - V1^2)/2*V1 = 1.01*10^5(0.012^2 - 0.006^2)/2*0.006 = 909 J

internal energy change of gas, U1 = nCv(T2 - T1)

Cv = 3R/2

U1 = 0.243*3*8.31(1200.398-300)/2 = 2727.301 J

process 2 -> V*T constnat = k2

P3 = 3.556 atm

T3 = ?

V3 = 9 lit = 0.009 m^3

using V2T2 = V3T3

T3 = V2T2/V3 = 0.012*1200.398/0.009 = 1600.5306 K

work done in process 2, dW = PdV

now, P = nRT/V

T = k2/V

P = nRk2/V^2

so, dW = nRk2dV/V^2

integrating

work done by gas, W2 = nR*k2*(1/V2 - 1/V3) = 0.243*8.31*0.009*1600.5306(1/0.012 - 1/0.009) = -807.999 J

internal energy change of gas, U2 = nCv(T3 - T2)

Cv = 3R/2

U2 = 0.243*3*8.31(1600.5306-1200.398)/2 = 1211.999 J

process 3 -> P*T = constant = k3

P4 = ?

T4 = 300 K

V4 = 0.3164 lit = 0.3164*10^-3 m^3

using P3T3 = P4T4

P4 = 3.556*1600.5306/300 = 18.971 atm = 19.16*10^5 Pa

work done in process 3, dW = PdV

now, P = k3/T

and PV = nRT

PV = nR(k3/P)

P^2 = k3*nR/V

P = sqroot(k3*nR)/sqroot(V)

so, dW = sqroot(k3*nR)dv/sqroot(V)

integrating

work done by gas, W3 = 2*sqroot(P4*T4*n*R)[sqroot(V4) - sqroot(V3)] = 2*sqroot(19.16*10^5*300*0.243*8.31)[sqroot(0.3164*10^-3) - sqroot(0.009)] = -5252.156 J

internal energy change of gas, U3 = nCv(T4 - T3)

Cv = 3R/2

U3 = 0.243*3*8.31(300 - 1600.5306)/2 = -3939.300684747 J

process 4 -> T = constant = 300 K

work done in process 4, dW = PdV

PV = nRT = constant

P = nRT/V

so, dW = nRTdV/V

integrating

work done by gas, W4 = nRT4*ln(V1/V4) = 0.243*8.31*300*ln(6/0.3164) = 1782.5681 J

internal energy change of gas, U4 = nCv(T1 - T4) = 0 J

Step number Work done by system Work done on system Internal energy change

1 909 J -- 2727.301 J

2 --- 807.999 J 1211.999 J

3 --- 5252.156 J -3939.300 J

4 1782.5681 J -- 0 J

Net work done by the system = -3368.5869 J [ i.e. work is done on the system]

total internqal energy change = 0 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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