Answer to Question #90490 in Molecular Physics | Thermodynamics for bella

Question

Two moles of a monatomic ideal gas (ɣ = 1.67) initially at pressure P1, volume V1, temperature T1 undergoes an isobaric process to volume V2 = 2V1. The gas is expanded adiabatically to V3 = 3V1. Determine final temperature of gas T3 if T1 = 298 K, and sketch the diagram for these processes

Answer 1

Process 1-2 is isobaric. Hence,

$$\frac{V_1}{T_1} = \frac{V_2}{T_2} \, \Rightarrow T_2 = T_1 \frac{V_2}{V_1} = 2 T_1$$

Process 2-3 is adiabatic. Hence,

$$p_2 V_2^\gamma = p_3 V_3^\gamma \, \Rightarrow \, \frac{p_3}{p_2} = \left( \frac{V_2}{V_3} \right)^\gamma = \left( \frac{2}{3} \right)^\gamma$$

Finally, according to the Clapeyron-Mendeleev equation

$$\frac{p_2 V_2}{T_2} = \frac{p_3 V_3}{T_3} \, \Rightarrow \, T_3 = T_2 \frac{p_3}{p_2} \frac{V_3}{V_2} = 2 T_1 \left(\frac{2}{3}\right)^\gamma \frac{3}{2} = 2 T_1 \left(\frac{2}{3}\right)^{(\gamma-1)}$$

Substituting the numerical values, we obtain:

$$T_3 = 2 \cdot 298 \left(\frac{2}{3} \right)^{1.67-1} \approx 454 \, K$$

Sketch:

Answer: 454 K