In Fig. 20-25, where V23= 3.00 V1, n moles of a diatomic...

In Fig. $20-25,$ where $V_{23}=$ $3.00 V_{1}, n$ moles of a diatomic ideal gas are taken through the cycle with the molecules rotating but not oscillating. What are (a) $p_{2} / p_{1},$ (b) $p_{3} / p_{1}$, and (c) $T_{3} / T_{1} ?$ For path $1 \rightarrow 2,$ what are(d) $W / n R T_{1}$(e) $Q / n R T_{1}$(f) $\Delta E_{\text {int }} / n R T_{1},$ and $(\mathrm{g}) \Delta S / n R ?$ For path $2 \rightarrow 3,$ what are (h) $W / n R T_{1}$,(i) $Q / n R T_{1},(\mathrm{j}) \Delta E_{\text {int }} / n R T_{1},(\mathrm{k}) \Delta S / n R ?$For path $3 \rightarrow 1,$ what are (I) $W / n R T_{1}$$(\mathrm{m}) Q / n R T_{1}$(n) $\Delta E_{\text {int }} / n R T_{1},$ and(o) $\Delta S / n R ?$

In Fig. $20-25,$ where $V_{23}=$ $3.00 V_{1}, n$ moles of a diatomic ideal gas are taken through the cycle with the molecules rotating but not oscillating. What are (a) $p_{2} / p_{1},$ (b) $p_{3} / p_{1}$, and (c) $T_{3} / T_{1} ?$ For path $1 \rightarrow 2,$ what
are(d) $W / n R T_{1}$(e) $Q / n R T_{1}$(f) $\Delta E_{\text {int }} / n R T_{1},$ and $(\mathrm{g}) \Delta S / n R ?$ For path $2 \rightarrow 3,$ what are (h) $W / n R T_{1}$,(i) $Q / n R T_{1},(\mathrm{j}) \Delta E_{\text {int }} / n R T_{1},(\mathrm{k}) \Delta S / n R ?$For path $3 \rightarrow 1,$ what are (I) $W / n R T_{1}$$(\mathrm{m}) Q / n R T_{1}$(n) $\Delta E_{\text {int }} / n R T_{1},$ and(o) $\Delta S / n R ?$

Key Concepts

Entropy Change in Processes

Entropy (S) is a measure of the disorder or randomness in a system, and its change during a process is determined by the heat exchanged divided by the temperature at which the transfer occurs (?S = ?dQ/T for reversible processes). Calculating entropy changes for various paths in a cycle helps illuminate the second law of thermodynamics, revealing the irreversibility and efficiency of the process.

Work, Heat, and Internal Energy in Processes

Each thermodynamic process (such as constant volume, constant pressure, or other paths) has specific ways in which work is done and heat is transferred. Work may be calculated as the integral of pressure with respect to volume (W = ?pdV), while changes in internal energy depend on the degrees of freedom of the gas. Understanding these relationships is crucial for determining the energy distribution in different parts of the cycle.

First Law of Thermodynamics

The First Law of Thermodynamics states that the change in a system's internal energy (?E_int) is equal to the heat (Q) added to the system minus the work (W) done by the system, expressed as ?E_int = Q - W. This law is a cornerstone in analyzing the energy balance along each process path in a cycle.

Degrees of Freedom in Diatomic Gases

Diatomic gases, depending on the conditions, have translational and rotational degrees of freedom that contribute to their internal energy. In problems where the molecules are rotating but not oscillating (i.e., no vibrational modes active), the gas typically exhibits five degrees of freedom, leading to an internal energy proportional to (5/2)nRT. This concept is essential for correctly accounting for energy changes during thermodynamic processes.

Ideal Gas Law

The Ideal Gas Law, expressed as pV = nRT, relates the pressure (p), volume (V), and temperature (T) of a gas with the number of moles (n) and the universal gas constant (R). This fundamental relationship is used to determine the state of a gas under different conditions and is key to analyzing any thermodynamic cycle involving gases.

Thermodynamic Cycles

A thermodynamic cycle is a series of processes through which a system returns to its initial state, meaning that the net change in internal energy over one complete cycle is zero. Analyzing such cycles requires calculating work done, heat transfer, and changes in entropy along each segment of the cycle, while ensuring consistency with the overall energy conservation principles.

Transcript

00:01 For this problem on the topic of entropy, we are shown a figure in which v23 is equal to 3 v1.

00:07 The number of moles in of a diatomic ideal gas are taken through the cycle with the molecules rotating but not oscillating.

00:16 We want to find p2 over p1, p3 over p1, t3 over t1.

00:21 And then for path 1 -2, we want to find w over nr t1, q times the log of r t1, delta e, the change in internal energy over.

00:30 Nr t 1 the entropy s times the natural log of r and then find the same for parts 2 3 and part 3 1.

00:40 Now the connection between molar heat capacity and degrees of freedom of a diatomic gas is given by setting f is equal to 5 in equation 51 which gives the molar heat capacity at constant volume c v to be 5r over 2 and the molar heat capacity at constant pressure cp to be 7r over 2 with gamma equal to 7 over 5.

01:06 Now the gas law in ratio form is used to obtain p2 and we get p2 is equal to p1 into the ratio of volumes v1 over v2 and this is equal to p1 divided by 3 which means that the ratio of pressures that we want p2 to p1 is 1 over 3 or 0.

01:34 3 .33.

01:42 For part b, the adiabatic relations equation 54 and 56 lead to p3 equal to p1 into v1 over v3, all to the power gamma, which is p1 divided by 3 to the power 1 .4, which implies that p3 over p1 is equal to 1 .4, which implies that p3 over p1 is equal to 1 over, 3 to the power 1 .4, which gives the ratios of pressure, the ratio of pressures to be 0 .215.

02:27 For part c, we can similarly find the temperature, t3, as t1, into v1 by v3 to the power gamma minus 1, which is t1 over 3 to the power 0 .4, which means that the power, that the ratio of temperatures that we want t3 over t1 is 1 over 3 to the power 0 .4 or 0 .644.

03:05 Next, for process 1, 2, we have for part d, the work given by the equation nrt1 times the log of v2 over v1, which is r -t -1 times the log of 3, which is 1 .1 r -t -1.

03:33 And so the ratio that we require, w over n -r -t -1 is equal to the log of 3, which is 1 .1.

03:49 For part e, the internal energy change delta e, internal, equal to zero since this is an ideal gas process without a temperature change.

04:01 Thus the energy absorbed as heat is given by the first law of thermodynamics.

04:08 And for part f, the internal energy change.

04:18 And so we've stated the internal energy change and we need to find the heat transfer for part e...

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