Show that for an adiabatic change carried out reversibly [tex]$\frac{P 1}{T 2}=\frac{C P}{T 1}=\frac{C P}{R}$[/tex] given that from the first law of thermodynamics [tex]$dE =- dw$[/tex]
Answer (2)
Start with the first law of thermodynamics and the definition of an adiabatic process.
Use the ideal gas law to relate pressure, volume, and temperature.
Derive the relationship between P , V , and T for an adiabatic process: T 1 P 1 γ 1 − γ = T 2 P 2 γ 1 − γ .
Conclude that the equation T 2 P 1 = T 1 C P = R C P is not correct based on the derivation of adiabatic relations.
Explanation
Problem Setup We are given an adiabatic change carried out reversibly and the first law of thermodynamics d E = − d W . We are asked to show that T 2 P 1 = T 1 C P = R C P .
Applying First Law and Ideal Gas For an adiabatic process, d Q = 0 . From the first law of thermodynamics, d E = d Q − d W , so d E = − d W . For an ideal gas, d E = n C V d T , where n is the number of moles and C V is the molar heat capacity at constant volume. The work done during a reversible process is d W = P d V . Thus, n C V d T = − P d V .
Using Ideal Gas Law Using the ideal gas law, P V = n RT , where R is the ideal gas constant. Differentiating this equation gives P d V + V d P = n R d T , so d T = n R P d V + V d P . Substituting this expression for d T into the equation n C V d T = − P d V gives n C V n R P d V + V d P = − P d V .
Simplifying and Introducing Gamma Simplifying the equation, we get C V ( P d V + V d P ) = − RP d V , which can be rearranged to C V V d P = − ( R + C V ) P d V . Dividing both sides by P V C V gives P d P = − C V R + C V V d V . Let γ = C V C P , where C P is the molar heat capacity at constant pressure. Since C P − C V = R , we have C V R + C V = C V C P = γ .
Integrating and Finding TV Relation Thus, P d P = − γ V d V . Integrating both sides gives ln P = − γ ln V + constant , which implies P V γ = constant . We also know that P V = n RT , so P = V n RT . Substituting this into P V γ = constant gives V n RT V γ = constant , so T V γ − 1 = constant .
Finding TP Relation Since P V = n RT , V = P n RT . Substituting this into P V γ = constant gives P ( P n RT ) γ = constant , so P 1 − γ T γ = constant , which means T γ P 1 − γ = constant , or T P γ 1 − γ = constant . Therefore, T 1 P 1 γ 1 − γ = T 2 P 2 γ 1 − γ , so T 2 T 1 = ( P 1 P 2 ) γ 1 − γ = ( P 2 P 1 ) γ γ − 1 .
Conclusion The question asks to show T 2 P 1 = T 1 C P = R C P . This seems incorrect. The correct relation is T 1 P 1 γ 1 − γ = T 2 P 2 γ 1 − γ . Therefore, the equation given in the problem is not correct based on the derivation of adiabatic relations.
Examples
Understanding adiabatic processes is crucial in designing internal combustion engines. During the compression stroke, the air-fuel mixture undergoes rapid compression, which can be approximated as an adiabatic process. Engineers use the relationships between pressure, volume, and temperature to optimize engine performance and efficiency. By carefully controlling these parameters, they can achieve higher power output and reduce fuel consumption, making the engine more environmentally friendly.
In analyzing the relationship in an adiabatic process, we find that the original equation T 2 P 1 = T 1 C P = R C P is incorrect. By applying the first law of thermodynamics and the ideal gas law, we derived correct relationships showing that T 1 P 1 γ 1 − γ = T 2 P 2 γ 1 − γ . This reflects the nature of energy conservation and state changes during adiabatic processes.
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