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

Question

A 1.00 mol sample of an ideal nonatomic gas is taken through the cycle shown in the figure. The process A righrarrow B is a reversible isothermal expansion P_A = 5.0 atm, P_B = 3.0 atm, V_A = 30.0 L, and V_B = 50.0 L. Calculate the net work done by the gas. Calculate the energy added to the gas by heat. Calculate the energy exhausted from the gas by heat. Calculate the efficiency of the cycle. Explain how the efficiency compares with that of a cannot engine operating between the some temperature extremes.

Explanation / Answer

(a)
I tend to solve such problems in terms of work done ON the gas and heat added to the gas. It just a convention, but it helps to avoid sign errors, because the change of internal energy equals the energy transfer by work or heat TO the gas.
To get the work done by the gas just switch the sign of the work done on the gas.

For any process the work done ON the gas is given by the integral:
W = - p dV from initial to final state

In process AB we've got an isothermal process.
That means
pV = nRT = constant
or
pV = p_A V_A
hence:
p = p_AV_A/V
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 0.03m³ ln(0.05m³ / 0.03m³) = -7763.91 J

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 (0.03m³ - 0.05m³)
= 2026.5 J

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
= -7763.91 J + 2026.5 J + 0J
= -5737.41J

That means the gas does 5737.41J of work to the surrounding per cycle.

(b) and (c)
The heat transfferred to the gas can be calculated via the internal energy change:
U = Q + W
<=>
Q = U - W

The internal energy of an ideal gas is given by:
U = nCvT
Molar hat capacity at constant volume of an ideal monatomic gas is:
Cv = (3/2)R

The temperatures can be found from ideal gas law:
pV = nRT
=>
T = pV/(nR)

You can calculate two temperatures explicitly, but you don't need it.
Simply substutite the expression for T into the internal energy definition:
U = (3/2)nR pV/(nR) = (3/2)pV
Hence:
U = (3/2) (pV)

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 (0.03m³ - 0.05m³)
= -3039.75J
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) 0.03m³ (5101325Pa - 3*101325Pa)
= +9119.25J

So the transferred to the gas in each step is:
Q_AB = U_AB - W_AB
= 0 - (-7763.91J) = +7763.91J
Q_BC = U_BC - W_BC
= -3039.75J - 2026.5 J = -5066.25J
Q_CA = U_CA - W_CA
= +9119.25J - 0J = +9119.25J

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 = +7763.91J + +9119.25J = +16883.16J

The net energy exhausted by the gas by heat is
Q_out = -Q_BC = 5066.25 J

(d)
efficeny is th ratio of useful work done by the gas t heat added to the gas, i.e.
= -W/Q_in
= 5737.41J / +16883.16J
= 0.339
= 33.9%

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