A 1.00-mol sample of an ideal monatomic gas is taken through the cycle shown in

Question

A 1.00-mol sample of an ideal monatomic gas is taken through the cycle shown in the figure. The process A ? B is a reversible isothermal expansion where PA = 5.0 atm, PB = 2.0 atm, VA = 20.0 L, and VB = 50.0 L.

(a) Calculate the net work done by the gas.
3.2 kJ

(b) Calculate the energy added to the gas by heat.
________ kJ

(c) Calculate the energy exhausted from the gas by heat.
________ kJ

(d) Calculate the efficiency of the cycle.
________ %

(e) Explain how the efficiency compares with that of a Carnot engine operating between the same temperature extremes.

Explanation / Answer

So the work done in the process is
W_AB = - p_AV_A/V dV dV from V_A to V_B
= - p_AV_A 1//V dV dV from V_A to V_B
= - p_AV_A ln(V_B/V_A)
Convert to SI-units in order to get a result in Pa m³ =J
W_AB = - 5101325Pa 0.02m³ ln(0.05m³ / 0.02m³) = - 16307J

In process BC we've got a constant pressure process.
So the work integral simplifies to
W = - p dV = -pV
Hence:
W_BC = - p_B (V_C - V_B)
= - 101325Pa (0.02m³ - 0.05m³)
= 3039.75J

In process CA there is no volume change. So no work is done
W_CA = 0J

So the net work done one the gas in the whole cycle ABCA is
W = W_AB + W_BC + W_CA
= -16307J + 3039.75J + 0J
= -13267J

That means the gas does 13267 J of work to the surrounding per cycle

b,c)

The the internal energy change of the the processes is
U_AB = 0 because it is an isothermal proceass
U_BC = (3/2) (p_CV_C - p_BV_B)
(because p_C = p_B)
= (3/2) p_B (V_C - V_B)
= (3/2) 101325Pa (0.02m³ - 0.05m³)
= -4560J
U_CA = (3/2) (p_AV_A - p_CV_C)
(because V_C = V_A)
= (3/2) V_C (p_A - p_C)
= (3/2) 0.02m³ (5101325Pa - 101325Pa)
= +12159J

So the transferred to the gas in each step is:
Q_AB = U_AB - W_AB
= 0 - (-16307J) = +16307J
Q_BC = U_BC - W_BC
= -3039.75J - 4560J = -7599 J
Q_CA = U_CA - W_CA
= +4560J - 0J = +4560J

Negative sign indicates that energy is exhausted in process BC.

The net energy added to the gas by heat is
Q_in = Q_AB + Q_CA = 16307+4560 = +20,867J

The net energy exhausted by the gas by heat is
Q_out = -Q_BC = 7599 J

d)

efficeny is th ratio of useful work done by the gas t heat added to the gas, i.e.
= -W/Q_in

= 13267/20867 = 0.635

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