A Carnot cycle is operated with liquid-gas interface. The...

A Carnot cycle is operated with liquid-gas interface. The vapor pressure is $p_{\mathrm{v}}$, temperature $T$, volume $V$. The cycle is operated according to the following $p-V$ diagram. The cycle goes isothermally from 1 to 2 , evaporating $n$ moles of liquid. This is followed by reversible cooling from 2 to 3 , then there is an isothermal contraction from 3 to 4 , recondensing $n$ moles of liquid, and finally a reversible heating from 4 to 1 , completes the cycle. FIGURE CAN'T COPY. (a) Observe that $V_2-V_1=V_{\mathrm{g}}-V_{\ell}$ where $V_{\mathrm{g}}=$ volume of $n$ moles of gas, $V_{\ell}=$ volume of $n$ moles of liquid. Calculate the efficiency in terms of $\Delta p, V_{\mathrm{g}}-V_{\ell}$, and $L_{\mathrm{v}}=$ latent heat vaporization of a mole of liquid. Treat $\Delta p$ and $\Delta T$ as small. (b) Recognizing that any two Carnot engines operating between $T$ and $T-\Delta T$ must have the same efficiency (why?) and that this efficiency is a function of $T$ and $T$ alone, use the result of part (a) to obtain an expression for $d p_{\mathrm{v}} / d T$ in terms of $V_{\mathrm{g}}-V_{\ell}, n, L_{\mathrm{v}}$ and $T$.

A Carnot cycle is operated with liquid-gas interface. The vapor pressure is $p_{\mathrm{v}}$, temperature $T$, volume $V$. The cycle is operated according to the following $p-V$ diagram.

The cycle goes isothermally from 1 to 2 , evaporating $n$ moles of liquid. This is followed by reversible cooling from 2 to 3 , then there is an isothermal contraction from 3 to 4 , recondensing $n$ moles of liquid, and finally a reversible heating from 4 to 1 , completes the cycle.
FIGURE CAN'T COPY.
(a) Observe that $V_2-V_1=V_{\mathrm{g}}-V_{\ell}$ where $V_{\mathrm{g}}=$ volume of $n$ moles of gas, $V_{\ell}=$ volume of $n$ moles of liquid. Calculate the efficiency in terms of $\Delta p, V_{\mathrm{g}}-V_{\ell}$, and $L_{\mathrm{v}}=$ latent heat vaporization of a mole of liquid. Treat $\Delta p$ and $\Delta T$ as small.
(b) Recognizing that any two Carnot engines operating between $T$ and $T-\Delta T$ must have the same efficiency (why?) and that this efficiency is a function of $T$ and $T$ alone, use the result of part (a) to obtain an expression for $d p_{\mathrm{v}} / d T$ in terms of $V_{\mathrm{g}}-V_{\ell}, n, L_{\mathrm{v}}$ and $T$.

Key Concepts

Carnot Cycle

The Carnot cycle is an idealized thermodynamic cycle that represents the maximum possible efficiency a heat engine can achieve operating between two reservoirs. Its efficiency depends only on the absolute temperatures of the hot and cold reservoirs, which makes it fundamental for understanding the upper limits of any engine cycle.

Isothermal Process

An isothermal process occurs at constant temperature. In the context of the Carnot cycle, the heat transfer during the isothermal expansion or compression is precisely balanced by the work done, making it central for analyzing energy changes during phase transitions without any temperature change.

Latent Heat of Vaporization

Latent heat of vaporization is the amount of energy required to change a substance from its liquid phase to its gaseous phase without changing its temperature. This concept is critical when considering phase transitions in thermodynamic cycles, as it quantifies the energy input or output during the phase change at the vapor?liquid interface.

Reversible Process

A reversible process is an idealized thermodynamic process that happens infinitely slowly so that the system is always in thermodynamic equilibrium. This concept is important in the Carnot cycle because only reversible processes ensure the maximal efficiency and serve as a benchmark for real engines.

p-V Diagram

A pressure–volume (p-V) diagram graphically represents the changes in pressure and volume during a thermodynamic process. It is a valuable tool in visualizing and calculating the work done during each stage of a cycle, including the Carnot cycle, where the area under the curve corresponds to work performed by or on the system.

Clausius-Clapeyron Relation

The Clausius-Clapeyron relation provides a way to understand how the vapor pressure of a substance changes with temperature during a phase transition. It relates the rate of change of the vapor pressure with temperature to the latent heat of vaporization and the volume change between the two phases, hence linking macroscopic thermodynamic measurements with microscopic phase change energetics.

Transcript

00:01 This question we are doing calculations on the carnal cycle.

00:04 Okay, so we are given this cycle and the gas starts at point a, the pressure is at 1 ,400 kilo -pascal, temperature 720 kelvin, 10 liters of volume, and then it expands isothermally to point b, and then it expands adibatically to point b, and then it expands adiatically to point c where the volume is 24 liters and then it compresses isothermally to points d and the volume is 15 and from d to a it compresses adibatically okay so we need to determine the pressures volume and temperatures for points b c and d okay okay so to do this problem, we need these equations.

01:08 Okay, it will be using the ideal guess equation, pv equals to nrt.

01:19 We also be using the equation for a diabetic process, pv to the power of gamma, it goes to constant, and this is also another equation we can use this also t times v to the gamma minus 1.

01:39 It's also equal to constant.

01:41 Okay yeah so now we can actually find all the temperatures okay so ta equals to tb equals to 720 kelvin because it's our isothermal process okay and then we have you can also find t a va va to the gamma minus 1 equals to td, bd, gamma minus 1.

02:24 Okay, oh yes, and gamma is 5 over 3 because it's non -atomic.

02:35 Okay, so we have td it goes to t .a times va over bd, gamma minus 1.

02:49 So this is 720 times 10 devised by 15.

02:55 To the power of 2 3rd, this td is 549 kelvin, and then tc equals to td equals to 549 kelvin, a again isothermal, okay, right? so you have, so just tb equals to 720 kelvin, tc equals to 720 kelvin, tc goes to td, it goes to 549 kelvin.

03:37 So just some boxes, boxing up some answers for the temperatures.

03:46 Okay, after we obtain the temperatures, we can find pressures.

03:54 So from the igu gas equation, if you know that p times v divided by t equals to constant.

04:13 So, p, a, va divided by ta equals to pd, vd, divided by td, equals to pc, vc, divide by tc.

04:33 Okay, so we can find pd.

04:38 Apd equals to pa times td over ta times va over vd.

04:46 So you have 1 ,400 kilo pascal multiply by 549 divided by 720.

05:01 This is the temperature part and then the volume part is 10 or 15 and you calculate you get 700 to 712 kilo pascal and then you do the same thing for pc.

05:17 So pc goes to pd times vd over vc okay because same temperature so there's no need to include the temperature term so you have siren 712 kilo pascal times 15 divide by 24 so you get 445 kilo pascal okay then so just to write down some answers.

06:21 Okay, next we'll find the volume for points b.

06:26 Okay, so for points b, we'll be using kbb to the gamma minus 1 equals to tcvc gamma minus 1.

06:37 So we can find vb.

06:41 B.

06:44 Okay, so let's just write out this thing.

06:48 So this is maybe over vc to the power two third equals to tc divided by tb, okay, which is 549 divided by 720.

07:04 Okay, so we be is equal to vc which is 24 times 549 divided by 10 to the power 3 half.

07:17 So you calculate you get 16 liters.

07:22 Okay, now just use the original values.

07:25 You get 16 liters nicely.

07:28 Okay.

07:29 And then we are then to find pb.