# Checking my reasoning on a carnot cycle problem

• Emspak
In summary: R(P_h - P_c)}Now, we can use the fact that the adiabatic process is also reversible to relate the temperatures at the hot and cold reservoirs to the pressures using the equation of state. This gives us:\frac{T_h}{T_c} = \left(\frac{P_h}{P_c}\right)^{\gamma - 1}Substituting this into our efficiency equation and simplifying, we get:\eta = \frac{2c_v}{c_v + const/R(P_h - P_c)} = \frac{2c_v}{c_v + const/R(P_h/P_c)^{\gamma - 1}(P_c - P_c
Emspak

## Homework Statement

An approximate equation of state for a gas is $P(v-b) = RT$ with b constant. The specific internal energy of the gas is $u = c_vT + const$. a) show that the specific heat at constant pressure of this gas is $c_v +R$ and b) show that the equation of a reversible adabiatic process is $P(v-b)^{\gamma} = const$ and c) show that the efficiency of a carnot cycle using this gas as a working substance is the same as tht for an ideal gas assuming $\left(\frac{\partial u}{\partial v}\right)_T = 0$

2. The attempt at a solution

I did the following for part a: we know that $c_P - c_V = \left[ \left( \frac{\partial u}{\partial v} \right)_T + P \right] \left( \frac{\partial v}{\partial T} \right)_P$ so since $\left( \frac{\partial v}{\partial T} \right)_P = \frac{R}{P}$ and $\left( \frac{\partial u}{\partial v} \right)_T = 0$ we plug all that into the first equation and we end up with $c_P = R + c_V$

For part b) we know that in any reversible adabiatic process

$\left( \frac{\partial P}{\partial v}\right)_s = \frac{c_P}{c_V} \left( \frac{\partial P}{\partial v} \right)_T$

which means, since $\frac{c_P}{c_V} = \gamma$ and for an ideal gas $\frac{dP}{P} + \gamma \frac{dv}{v}$ then $ln P + \gamma ln v = 0$ and we end up with $Pv^{\gamma}$ which when we plug in our original expressions from the equation of state we have $\frac{RT}{(v-b)}\left(\frac{RT}{P} + b\right)^{\gamma}= const$ and that gets us $P(v-b)^{\gamma}$ with a little algebra.

So far so good, but the last part of the problem stumps me a bit and I think there's some really simple thing I am missing.

Here's the thing: I know the efficiency of a Carnot cycle is $\eta = \frac{T_2 - T_1}{T_2}$ and $\eta = \frac{Q_2 - Q_1}{Q_2}$ also. If I just plugged in T1 and T2 into the equation of state I suppose I might get an answer, and I wondered if I should just do that and the problem really is that simple. Or, given that they are asking us to assume something about the internal energy of the gas that there's some other silly step I should do. I know that this isn't that complicated so maybe I am just overthinking it. By inspection I suspect that doing the plug-in-T1 method I will get the same answer as for an ideal gas, since $T = \frac{P(v-b)}{R}$. But I wasn't sure if I should have a v1 or a P1 in there -- I was sure one would be left constant but I wasn't sure which one.

Anyhow, any help is appreciated.

Last edited:

To solve part c), you can use the fact that for a reversible adiabatic process, \Delta u = 0. This means that the change in internal energy during the process is equal to 0. Using the definition of specific internal energy given in the problem, we can rewrite this as:

c_v\Delta T + const = 0

Since the process is adiabatic, there is no heat transfer (Q = 0), so we can use the first law of thermodynamics to write:

\Delta u = Q - W = 0

where W is the work done by the gas during the process. Since the process is reversible, we can use the equation of state to relate the change in volume to the change in pressure:

\Delta v = \frac{RT}{P}\Delta P

Substituting this into the work equation and solving for \Delta T, we get:

\Delta T = -\frac{c_v}{R}\Delta P

Now, let's consider a Carnot cycle using this gas as the working substance. The efficiency of a Carnot cycle is given by:

\eta = \frac{W}{Q_h} = \frac{Q_h - Q_c}{Q_h} = 1 - \frac{Q_c}{Q_h}

where Q_h is the heat added to the system at the hot reservoir and Q_c is the heat removed from the system at the cold reservoir. Since the process is reversible, the heat added and removed can be written in terms of the change in internal energy:

Q_h = \Delta U_h = c_v\Delta T + const

Q_c = \Delta U_c = c_v\Delta T - const

Substituting these into the efficiency equation and using the expression we found for \Delta T, we get:

\eta = 1 - \frac{c_v\Delta T - const}{c_v\Delta T + const} = \frac{2c_v\Delta T}{c_v\Delta T + const}

Finally, substituting the expression for \Delta T that we found earlier, we get:

\eta = \frac{2c_v}{c_v + const/R\Delta P}

Using the fact that \Delta P = P_h - P_c (where P_h and P_c are the pressures at the hot and cold reservoirs, respectively), we can rewrite this as:

\eta = \frac

## 1. What is a Carnot cycle?

A Carnot cycle is a theoretical thermodynamic cycle that consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. It is used to model the most efficient heat engine possible, and serves as a benchmark for evaluating the performance of actual engines.

## 2. How do I determine the efficiency of a Carnot cycle?

The efficiency of a Carnot cycle can be determined using the Carnot efficiency formula: efficiency = 1 - (Tcold / Thot), where Tcold is the temperature of the cold reservoir and Thot is the temperature of the hot reservoir.

## 3. What is the purpose of checking my reasoning on a Carnot cycle problem?

Checking your reasoning on a Carnot cycle problem is important because it allows you to ensure that your calculations and assumptions are correct. This can help you identify any errors and improve your understanding of the concept.

## 4. What are the assumptions made in a Carnot cycle?

The assumptions made in a Carnot cycle include: the working fluid is ideal and follows the ideal gas law, all processes are reversible, there is no internal friction or heat transfer, and the system is in thermal equilibrium at all times.

## 5. How does a Carnot cycle relate to the real world?

While a Carnot cycle is a theoretical model, it serves as a benchmark for real-world heat engines. The closer an actual engine's performance is to the Carnot efficiency, the more efficient it is. Engineers use the Carnot cycle as a reference point to improve the efficiency of real-world engines.

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