A 4 liter sample of a diatomic gas with γ =1.4 confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00atm. and 300K. First, its pressure is tripled under constant volume. Then it expands adiabatically to its original pressure. Finally the gas is compressed isobarically to its original volume.
a) draw pV diagram of this cycle
b) Determine the volume of the end of the adiabatic expansion
c) Find the temperature of the gas at the start of the adiabatic expansion
d) Find the temperature at the end of the cycle
e) What was the net work done on the gas for this cycle
a) As a bot, I cannot physically draw a diagram, but I can describe it to you. The pV diagram for this cycle would have three steps:
1. The initial state is at 1.00atm and 300K. The gas is tripled in pressure under constant volume, so it moves vertically upwards.
2. The gas then expands adiabatically to its original pressure.
3. Finally, the gas is compressed isobarically back to its original volume, so it moves horizontally back to the starting point.
b) The volume at the end of the adiabatic expansion can be determined using the relation:
V2 = V1 * (P1/P2)^(1/γ), where V1 and P1 are the initial volume and pressure, and P2 is the pressure at the end of the adiabatic expansion.
c) To find the temperature at the start of the adiabatic expansion, we can use the relation:
T2 = T1 * (P2/P1)^((γ-1)/γ), where T1 is the initial temperature and T2 is the temperature at the start of the adiabatic expansion.
d) The temperature at the end of the cycle can be found using the ideal gas law equation:
P1 * V1 / T1 = P2 * V2 / T2, where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the pressure, volume, and temperature at the end of the cycle.
e) The net work done on the gas for this cycle can be calculated by the area enclosed by the pV diagram.
a) The pV diagram for this cycle can be divided into three steps:
Step 1: Pressure is tripled under constant volume. This is represented by a vertical line on the diagram, as the volume remains constant.
Step 2: The gas expands adiabatically to its original pressure. This is a diagonal line downwards on the diagram, representing the expansion of the gas.
Step 3: The gas is compressed isobarically to its original volume. This is represented by another vertical line on the diagram, as the volume remains constant.
Here is a representation of the pV diagram:
|
| Step 3
P |=================
| Step 2
| ===============
| Step 1
|
----------------- V
b) To determine the volume at the end of the adiabatic expansion, we can use the formula for adiabatic expansion:
(V1/V2)^(γ-1) = (P2/P1)^(1-γ)
Where V1 and P1 are the initial volume and pressure, and V2 and P2 are the final volume and pressure.
In this case, V1 = 4 liters, P1 = 1.00 atm, P2 = 1.00 atm (as the final pressure is the same as the initial pressure), and γ = 1.4.
Plugging in these values, we have:
(V1/V2)^(1.4-1) = (1.00/1.00)^(1-1.4)
Simplifying, we get:
(V1/V2) = (1.00/1.00)^(1/0.4)
V1/V2 = 1.585
V2 = V1 / 1.585
V2 = 4 liters / 1.585 = 2.52 liters
Therefore, the volume at the end of the adiabatic expansion is 2.52 liters.
c) To find the temperature at the start of the adiabatic expansion, we can use the adiabatic process equation:
T2/T1 = (V1/V2)^(γ-1)
Where T1 and T2 are the initial and final temperatures, and V1 and V2 are the initial and final volumes.
In this case, T1 = 300K, V1 = 4 liters, V2 = 2.52 liters, and γ = 1.4.
Plugging in the values, we have:
T2/300 = (4/2.52)^(1.4-1)
T2/300 = (1.587)^0.4
T2/300 = 1.118
T2 = 1.118 * 300
T2 = 335.4K
Therefore, the temperature at the start of the adiabatic expansion is 335.4K.
d) The temperature at the end of the cycle is the same as the initial temperature, as the gas is brought back to its original state.
Therefore, the temperature at the end of the cycle is 300K.
e) The net work done on the gas for this cycle can be determined using the first law of thermodynamics:
ΔU = Q - W
Where ΔU is the change in internal energy, Q is the heat added to the system, and W is the net work done on the system.
In this case, the gas is taken through a closed cycle, so the net change in internal energy is zero.
Therefore:
ΔU = 0 = Q - W
Solving for W, we find:
W = Q
Since the process is adiabatic, meaning no heat is transferred in or out of the system during the expansion, the heat added to the system is zero.
Therefore, the net work done on the gas for this cycle is also zero. The area enclosed by the pV diagram is a rectangle, and the work done in the clockwise direction is equal to the work done in the counterclockwise direction, resulting in a net work of zero.
a) To draw a pV diagram of the cycle, we need to plot the pressure (p) on the y-axis and the volume (V) on the x-axis. Based on the given information, the cycle consists of three processes:
1. Constant Volume Process: The pressure is tripled while the volume remains constant. This means that the initial and final volumes are the same. So, this process is represented by a vertical line on the pV diagram, starting from the initial pressure and ending at three times the initial pressure.
2. Adiabatic Expansion Process: The gas expands adiabatically, which means no heat is exchanged with the surroundings. The volume increases while the pressure decreases. The adiabatic expansion process is represented by a curve sloping downwards on the pV diagram since the pressure decreases as the volume increases.
3. Isobaric Compression Process: The gas is compressed isobarically, meaning the pressure remains constant. The gas is compressed back to its original volume. This process is represented by a horizontal line on the pV diagram, starting at the end of the adiabatic expansion curve and ending at the initial volume.
b) The volume of the gas at the end of the adiabatic expansion can be determined by finding the intersection point of the adiabatic expansion curve and the isobaric compression line in the pV diagram. The coordinates of this intersection point represent the volume at the end of the adiabatic expansion.
c) The temperature at the start of the adiabatic expansion can be found using the formula for adiabatic processes:
T1/T2 = (V2/V1)^(γ-1),
where T1 and T2 are the initial and final temperatures, V1 and V2 are the initial and final volumes, and γ is the specific heat ratio (given as 1.4).
d) The temperature at the end of the cycle can be found by using the ideal gas law:
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. Since the initial and final volumes are the same, you can use this equation to find the temperature at the end of the cycle.
e) The net work done on the gas for this cycle can be determined by calculating the area enclosed by the pV diagram. Since work is defined as the area under the pV curve, you can calculate the areas corresponding to each process (constant volume, adiabatic expansion, and isobaric compression) and then subtract the areas enclosed during expansion from the area enclosed during compression to find the net work done.