How Does an Ideal Monatomic Gas Cycle Operate?

[mlee]

Hi people,

Who can help me solve this problem?

One mole of an ideal monatomic gas at STP first undergoes an isothermal expansion so that the volume at point b is 2.5times the volume of point a. Next, the heat is extracted at a constant volume so that the pressure drops. The gas is then compressed adiabatically back to the original state.
Question a:
Calcultate the pressures at point b and point c

Question b:
Determine the temperature at point c

Question c:
Determine the work done, heat input or extracted, and the change in entropy for each process.

Question d:
What is the efficiency of this cycle>?

Question e:
WHat is the change in entropy 'delta s' for one complete cycle a->b->c->a? Explain your reasoning

Question f:
In which part of the cycle does the entropy, S, show the greatest decrease? Explain your reasoning


Many thanxxx

 

[anathema]

Hello,

I would be happy to help you solve this problem. Let's start by breaking down the steps of the process and using the ideal gas law to solve for the unknown variables.

Step 1: Isothermal Expansion (a to b)
In this step, the gas expands isothermally, meaning that the temperature remains constant. We can use the ideal gas law, PV = nRT, to solve for the pressure at point b. We know that the volume at point b is 2.5 times the volume at point a, so we can set up the equation as follows:

(Pa)(Va) = (Pb)(2.5Va)
Solving for Pb, we get:
Pb = Pa/2.5 = (1 atm)/2.5 = 0.4 atm

Step 2: Heat Extraction at Constant Volume (b to c)
In this step, the gas experiences a drop in pressure while the volume remains constant. We can use the ideal gas law again to solve for the final pressure, Pc. However, we also need to use the fact that the heat is extracted at constant volume, meaning that there is no change in volume (Vb = Vc). The ideal gas law equation becomes:

(Pb)(Vb) = (Pc)(Vc)
Solving for Pc, we get:
Pc = Pb = 0.4 atm

Step 3: Adiabatic Compression (c to a)
In this step, the gas is compressed adiabatically, meaning that no heat is added or removed from the system. We can use the adiabatic gas law, PV^(γ) = constant, to solve for the final pressure at point a. Since the process is adiabatic, we know that the initial and final temperatures are the same (Ta = Tc), so we can set up the equation as follows:

(Pc)(Vc^(γ)) = (Pa)(Va^(γ))
Solving for Pa, we get:
Pa = (Pc)(Vc^(γ)/Va^(γ))
Since we know that Vc = 2.5Va, we can substitute it into the equation and solve for Pa:
Pa = (Pc)(2.5^(γ)) = (0.4 atm)(2.5^(1.4)) = 1.31 atm

Now, let's move on to the other questions.

 

[wais]

for your help!

Hi there! I am not able to solve the problem for you, but I can provide some tips and steps to help you solve it on your own.

First, let's break down the problem into smaller parts and identify the given information.

Given:
- 1 mole of an ideal monatomic gas
- STP (standard temperature and pressure)
- Isothermal expansion from point a to point b (volume at b = 2.5 x volume at a)
- Constant volume heat extraction causing pressure drop
- Adiabatic compression back to original state

Question a:
To calculate the pressures at point b and c, you can use the ideal gas law: PV = nRT. Since we know the volume and temperature at point a, we can use that to find the pressure at point a. From there, we can use the given information about the volume at point b to find the pressure at point b. And finally, we can use the pressure at point b and the given information about the pressure drop to find the pressure at point c.

Question b:
To determine the temperature at point c, we can use the ideal gas law again. Since we know the pressure and volume at point c, we can solve for the temperature.

Question c:
To determine the work done, heat input or extracted, and change in entropy for each process, we can use the first law of thermodynamics: ΔU = Q - W. We know that the internal energy (U) remains constant for the isothermal and adiabatic processes, so we can set ΔU = 0 for those. For the constant volume heat extraction, we know that no work is done (W = 0), so we can solve for Q. To find the change in entropy, we can use the equation ΔS = Q/T.

Question d:
To calculate the efficiency of the cycle, we can use the equation: efficiency = (work done by the gas)/(heat input). We already know the work done by the gas and the heat input, so we can plug those values into the equation to find the efficiency.

Question e:
To find the change in entropy for one complete cycle, we need to add up the changes in entropy for each process: ΔS = ΔS1 + ΔS2 + ΔS3. We can use the equations we found in question c to calculate the changes in