Question

Rectangular PV Cycle 1 2 3 4 A piston contains 600 moles of an ideal monatomic gas that initally has a pressure of 2.31 x 10 Pa and a volume of 3.8 m'. The piston is connected to a hot and cold reservoir and the gas goes through the following quasi-static cycle accepting energy from the hot reservoir and exhausting energy into the cold reservoir. 1. The pressure of the gas is increased to 5.31 x 109 Pa while maintaining a constant volume. 2. The volume of the gas is increased to 10.8 m while maintaining a constant pressure. 3. The pressure of the gas is decreased to 2.31 x 10 Pa while maintaining a constant volume. 4. The volume of the gas is decreased to 3.8 m while maintaining a constant pressure. It may help you to recall that Cy - 12.47 J/K/mole and Cp - 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022102) times the number of moles of the gas. 1) How much energy is transferred into the gas from the hot reservoir? J Submit 2) How much energy is transferred out of the gas into the cold reservoir? 3) How much work is done by the gas? Submit 4) What is the efficiency of this cycle?

Answer 1

a. How much energy is transferred into the gas from the hot reservoir?

Step 1 transfers heat to the system so we can use the Constant Volume Equation of:

heatEnergy = specific heat constant of volume * initial Volume * (final pressure - initial pressure) / R constant

=

12.47J/K *mole * 3.8m** (5.31 – 2.31) * 105 - = 1732429.91 8.2057m'atm/molK

Step 2 requires another additional heat source in order to maintain the pressure at a constant as the volume expands. We use the system of:

heatEnergy =  specific heat constant of pressure * pressure * (final volume - initial volume) * (1.0 / R constant)

20.79J/K * mole* (10.8 – 3.8)m** 2.31 * 105 - = 4096838.7828J 8.2057m'atm/mol K

Combining both of these energies produces the total
heat energy entering the system =5829268.6828 J

b. How much energy is transferred out of the gas into the cold reservoir?

From the last problem we know that a certain amount of heat entered the reservoir. The next two energy processes lead to a decrement in the heat of the system and eventually produce the amount of heat actually lost in the system. We basically do the same as last question but just move in reverse.

Step 3 requires a reduction of heat from the system by maintaining a constant volume while reducing the pressure of the system and transferring energy to the reservoir

heatEnergy =
specific heat constant of volume * secondVolume *
(final pressure - initial pressure) / R constant

12.47J/K * mole* 10.8m** (5.31 – 2.31) * 105 8.2057m'atm/molk = 4923748.13684)

Step 4 requires a reduction in volume while maintaing a cosntant pressure and transferring energy to cold reservoir

heatEnergy =
specific heat constant of pressure * pressure *
(final volume - initial volume) * (1.0 / R constant)

20.79J/K *mole* (10.8 - 3.8)m** 5.31 * 105 - = 9417408.63059 8.2057m'atm/molK

Add both negative values from step 3 & 4, then negate since it only
asks for the amount of energy that has moved
= 14341156.7674 J

c. How much work is done by the gas?

The amount of work done by the gas is the difference between
both the amount of heat added to the system and subtracted
the amount of heat that has left the system.
Work Done By Gas
= heat entered - heat exited
= 5829268.6828 - 14341156.7674 = - 8511888.08463 J

d. What is the efficiency of this cycle?

Efficiency of a cycle is defined by the amount of work
done the gas divided by the total amount of energy that
has entered the system to create a ratio of Work/Heat Added.
Efficiency = Work / (Heat Added)
= 8511888.08463 / 14341156.7674
= 0.5935