To find the efficiency of the engine, we need to first calculate the amount of work done by the engine during the cycle and the amount of heat supplied to the engine. The efficiency is then given by the ratio of the work done to the heat supplied.
1) Isothermal expansion: During this step, the gas expands from a volume of 18.7 L to 33 L at a constant temperature of 370 K. Since the process is isothermal, the gas obeys the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.
The pressure of the gas during this step can be found by rearranging the ideal gas law:
P = nRT/V
Substituting the given values, we get:
P = (1 mol)(8.314 J/mol/K)(370 K)/(18.7 L) = 162.7 kPa
The work done during this step is given by:
W1 = -∫PdV
where the integral is taken from the initial volume of 18.7 L to the final volume of 33 L. Since the process is isothermal, the pressure can be substituted into the integral:
W1 = -nRT ln(Vf/Vi)
Substituting the given values, we get:
W1 = -(1 mol)(8.314 J/mol/K)(370 K) ln(33 L/18.7 L) = -8201 J
2) Cooling at constant volume: During this step, the gas is cooled from a temperature of 370 K to 209 K at a constant volume of 18.7 L. Since the volume is constant, no work is done during this step. The amount of heat removed from the gas is given by:
Q2 = nCΔT
where C is the heat capacity and ΔT is the change in temperature. Substituting the given values, we get:
Q2 = (1 mol)(21 J/K)(209 K - 370 K) = -3366 J
Note that the negative sign indicates that heat is being removed from the gas.
3) Isothermal compression: During this step, the gas is compressed back to its original volume of 18.7 L at a constant temperature of 209 K. Using the same approach as in step 1, we can find the pressure of the gas:
P = (1 mol)(8.314 J/mol/K)(209 K)/(18.7 L) = 93.0 kPa
The work done during this step is given by:
W3 = -nRT ln(Vi/Vf) = (1 mol)(8.314 J/mol/K)(209 K) ln(18.7 L/33 L) = 4082 J
Note that the negative sign indicates that work is being done on the gas.
4) Heating at constant volume: During this step, the gas is heated back to its original temperature of 370 K at a constant volume of 18.7 L. Again, no work is done during this step. The amount of heat added to the gas is given by:
Q4 = nCΔT = (1 mol)(21 J/K)(370 K - 209 K) = 3381 J
The net work done by the engine during the cycle is given by:
Wnet = W1 + W3 = -8201 J + 4082 J = -4120 J
The net heat supplied to the engine during the cycle is given by:
Qnet = Q2 + Q4 = -3366 J + 3381 J = 15 J
The efficiency of the engine is given by:
η = Wnet/Qnet = -4120 J/15 J = -275.3
Note that the negative sign indicates that the engine is not operating as a heat engine, but rather as a heat pump or refrigerator. This is because the amount of work done by the engine is greater than the amount of heat supplied to the engine, which violates the second law of thermodynamics.