The pressure-volume graph shows an engine cycle where the working substance is an ideal monatomic gas (γ = 5/3). The process A-B is adiabatic. At point A, the pressure is 2.70×105 Pa and the volume is 0.0180 m3. At point B, the volume is 0.0450 m3. During a cycle, calculate (a) the heat added to the gas, (b) the heat removed from the gas (stated as a positive number), and (c) the net work done.
THE GRAPH looks like
P
A
^
C < B
V
To calculate the heat added to the gas (Qin), you can use the first law of thermodynamics:
Qin = ΔU + W
where Qin is the heat added to the gas, ΔU is the change in internal energy of the gas, and W is the work done on the gas.
In an ideal gas, the internal energy is given by:
ΔU = (3/2) * n * R * ΔT
where n is the number of moles of gas, R is the ideal gas constant, and ΔT is the change in temperature of the gas.
Since the process A-B is adiabatic, there is no heat transfer (Q = 0). Therefore, the heat added to the gas (Qin) is also zero.
(a) Qin = 0
To calculate the heat removed from the gas (Qout), you can use the same equation as above:
Qout = ΔU + W
In process B-C, the volume increases and the gas expands adiabatically. Therefore, no heat transfer occurs in this process (Q = 0). Hence, the heat removed from the gas (Qout) is also zero.
(b) Qout = 0
The net work done (W) on the gas can be calculated using the area enclosed by the pressure-volume graph. Since the process A-B is adiabatic, the net work done can be calculated using the formula:
W = (3/2) * n * R * (Tb - Ta)
where n is the number of moles of gas, R is the ideal gas constant, and Tb and Ta are the temperatures at points B and A, respectively.
Using the ideal gas law, we can relate pressure, volume, and temperature:
PV = nRT
From point A, we have:
P1 * V1 = n * R * T1
where P1 is the pressure at point A and V1 is the volume at point A.
Similarly, from point B, we have:
P2 * V2 = n * R * T2
where P2 is the pressure at point B and V2 is the volume at point B.
By rearranging these equations, we can solve for the temperatures T1 and T2:
T1 = P1 * V1 / (n * R)
T2 = P2 * V2 / (n * R)
Then, the net work done (W) is given by:
W = (3/2) * n * R * (T2 - T1)
(c) W = (3/2) * n * R * (T2 - T1)
Note: To obtain precise numerical values for (a), (b), and (c), you would need to substitute the given values of pressure, volume, and constants into the equations.
To find the heat added to the gas during the cycle, you need to use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. The equation can be written as:
ΔU = Q - W
Let's break down the problem into different parts:
(a) Finding the heat added to the gas:
Since the process A-B is adiabatic, no heat is transferred during this process. Therefore, the heat added to the gas during the cycle can be calculated by considering the entire cycle. In a closed cycle, the change in internal energy is zero, so ΔU = 0.
Therefore, Q = W.
(b) Finding the heat removed from the gas:
To find the heat removed from the gas, we need to consider the entire cycle. As mentioned before, ΔU = 0 for a closed cycle. So, Q - W = 0. Since Q = W, the heat removed from the gas is also equal to the work done by the gas during the cycle.
(c) Finding the net work done:
The net work done can be calculated by finding the area enclosed by the cycle on the pressure-volume graph. In this case, the cycle consists of two processes: A-B (adiabatic) and B-C (isochoric). The area under the curve representing process A-B represents the work done during this process. Similarly, the area under the curve representing process B-C represents the work done during this process.
Since process A-B is adiabatic, we can use the formula for adiabatic processes:
W = (γ / (γ - 1)) * (P2 * V2 - P1 * V1)
Where γ is the specific heat ratio of the gas, P1 and V1 are the initial pressure and volume (point A), and P2 and V2 are the final pressure and volume (point B).
To find the work done during process B-C, we need to know the pressure at point C, which is not given in the problem statement. Without this information, we cannot accurately calculate the net work done.
Therefore, without the pressure at point C, we can only calculate the heat added to the gas during the cycle and the heat removed from the gas. The net work done cannot be determined without additional information.