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Suppose we have a heat pump with 1 mol of some sort of fuel that goes through

  1. an isochoric process that takes the pressure of the gas from $P_2$ to $P_1$ while staying at $V_1$,

  2. an isobaric expansion that takes it from $V_1$ to $V_2$ at $P_1$

  3. an adiabatic compression that takes it from $(P_1,V_2)$ to $(P_2,V_1)$

question 1: What would be the temperature at each point? I am always confused when it comes to temperature, doesn't $T=\frac{PV}{NR}$ satisfy every scenario (in a case where specific values of P and V are given?

question 2: And is it correct that with the isochoric process the gas gives off heat as energy and the surrounding would be hotter and with the isobaric expansion the system is absorbing energy from the surrounding so the surrounding is getting cooler?

PS: in the adiabatic compression the internal energy and temperature also increase, but there is no heat flow, so the surrounding is not cooled, is this correct?

I just learned about adiabatic process and in the book it mentioned this is how a heat pump works, just reverse the process, so im just trying to see if i understood it.

EDIT:

Isochoric part: $\Delta Q=\Delta U +\Delta W$, since work done is defined to be $w=\int pdV$ and isochoric means volume stays constant so work done on the surrounding is 0. Then $\Delta Q=\Delta U$. Since pressure in this process is lowered, the volume stays the same, the internal energy is lowered because there is more room for the particles to move around, i.e. potential energy is lowered. The system's temperature decreased.

Isobaric part: As shown above, we know now there is work done, since its an expansion, work is done on the surroundings, then by the equation $\Delta Q=\Delta U +\Delta W$, heat is added to the system, so the surrounding's temperature lowered.

Adiabatic Part by definition no heat flows into or out of a system so this process has no effect on its surroundings.

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  • $\begingroup$ Let's see your analysis (algebraically) for the process you described. $\endgroup$ Jun 19, 2022 at 11:03
  • $\begingroup$ Hi I ve added some thoughts in, could you have a look? And if you could, some information to my question 1 would be greatly apprecitated $\endgroup$ Jun 19, 2022 at 13:13
  • $\begingroup$ Don't use $\Delta$ in front of $Q$ and $W$ for heat and work. Heat and work are not properties that change, like internal energy. $\endgroup$
    – Bob D
    Jun 19, 2022 at 15:18
  • $\begingroup$ The first step choice would be to solve the adiabatic compression first, for a specified pressure ratio or specified volume ratio. This will give the temperature after compression. $\endgroup$ Jun 19, 2022 at 18:00
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    $\begingroup$ Ah are you refering to question one, temperature for point 3, the adiabatic compression? Ah the equation is $T_2V_{2}^{\gamma -1}=T_3V_{3}^{\gamma -1}$ so we can use the temperature from the second state, so $T_3=T_2(\frac{V_2}{V_3})^{\gamma -1}$ $\endgroup$ Jun 19, 2022 at 18:28

1 Answer 1

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question 1: What would be the temperature at each point? I am always confused when it comes to temperature, doesn't $T=\frac{PV}{NR}$ satisfy every scenario (in a case where specific values of P and V are given?

You can use the equation $T=\frac{PV}{NR}$ for the relationship between pressure, volume and temperature for every equilibrium state in every scenario, provided the "fuel" is an ideal gas, because this is the equation of state for an ideal gas only.

question 2: And is it correct that with the isochoric process the gas gives off heat as energy and the surrounding would be hotter and with the isobaric expansion the system is absorbing energy from the surrounding so the surrounding is getting cooler?

For the isochoric process, the temperature of the surroundings is lower at the end of the process than at the beginning of the process, and for the isobaric expansion process the temperature of the surroundings at the end of the process is higher than at the beginning of the process.

PS: in the adiabatic compression the internal energy and temperature also increase, but there is no heat flow, so the surrounding is not cooled, is this correct?

That is correct.

Isochoric part:

Then $\Delta Q=\Delta U$. ....Since pressure in this process is lowered, the volume stays the same, the internal energy is lowered because there is more room for the particles to move around, i.e. potential energy is lowered.

The potential energy is not lowered. It stays the same. Only kinetic energy is lowered. That's because the change in internal energy of an ideal gas depends only on temperature change due to a change in the average kinetic energy of the particles, according to $\Delta U=NC_{V}\Delta T$.

Isobaric part: As shown above, we know now there is work done, since its an expansion, work is done on the surroundings, then by the equation $\Delta Q=\Delta U +\Delta W$, heat is added to the system, so the surrounding's temperature lowered.

As I already indicated above, the temperature of the surroundings at the end of the isobaric process is greater than the temperature of the surroundings at the beginning of the process.

Adiabatic Part by definition no heat flows into or out of a system so this process has no effect on its surroundings.

Your are correct there are no heat flows into or out of the system. But it's not correct to say the adiabatic process has no effect on the surroundings. For an adiabatic compression process the surroundings loses energy equal to the increase in internal energy of the system because of the work the surroundings does on the system.

Hope this helps.

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