Question

Consider an engine in which the working substance is 1.23 mol of an ideal gas for which γ = 1.41. The engine runs reversibly in the cycle shown on the PVdiagram (see figure below). The cycle consists of an isobaric (constant pressure) expansion a at a pressure of 15.0 atm, during which the temperature of the gas increases from 300 K to 600 K, followed by an isothermal expansion b until its pressure becomes 3.00 atm. Next is an isobaric compression c at a pressure of 3.00 atm, during which the temperature decreases from 600 K to 300 K, followed by an isothermal compression d until its pressure returns to15 atm. Find the work done by the gas, the heat absorbed by the gas, and the internal energy change of the gas, first for each part of the cycle and then for the complete cycle. Work done by the gas: Wa = kJ

Wb = kJ

Wc = kJ

Wd = kJ

Wtotal = kJ

Heat absorbed by the gas:

Qa = kJ

Qb = kJ

Qc = kJ

Qd = kJ

Qtotal = kJ

Internal energy change of the gas:

ΔUa = kJ

ΔUb = kJ

ΔUc = kJ

ΔUd = kJ

ΔUtotal = kJ

Answer 1

Phase a:

Work done in phase a (constant pressure) is simply : -P(V - V')

and V = n*R*T/P

therefore

Wa = -P[(n*R*T1/P )- (n*R*T2 /P)]

Wa = -1.23*8.31*(600-300) = -3066.40J

Negative sign because the gas is expanding.

Here the change in Internal Energy will be:

U = 3/2*n*R*T = 3/2*1.23*8.31*(600 - 300) = 4600 J

And so by First Law of Thermodynamic

U = Q + PV

we get the heat abosrbed to be: Q_a = 4600 + 3066.40 = 7666.40 J

Phase b:

Work done in phase b is given by:

Wb = n*R*T*ln(P1/P2) = 1.23*8.31*600ln(15/3) = -9870.33J

here the temperature is constant and so there is no change in internal energy.

and so the heat absorbed will be: Q_b = +9870.33 J

Phase c:

Work done in phase c will be:

Wc =

Wa = -P[(n*R*T1/P )- (n*R*T2 /P)]

Wa = -1.23*8.31*(300-600) = 3066.40J

U = 3/2*n*R*T = 3/2*1.23*8.31*(300 - 600) = -4600 J

and so the heat absorbed will be: -4600 - 3066.40 = -7666.40 J

Phase d:

finally the work done in phase d will be:

Wb = n*R*T*ln(P1/P2) = 1.23*8.31*300ln(3/15) = 4935.16 J

here again the change in internal energy will be zero. Therefore the heat absorbed = -4935.16 J

therefore the total work done will be: -3066.40 + 3066.40 - 9870.33 + 4935.16 = -4935.17 J

total heat absorbed = 7666.40 - 7666.40 +9870.33 -4935.16 =  4935.17 J

total change in internal energy = 0 J